Motion in a Straight Line

The body can be considered approximately a point object when its size is negligible compared to the distance travelled and its rotational motion is not important.

(a) A railway carriage moving without jerks between two stations — Yes

The length of the carriage is very small compared to the distance between the stations, so it can be treated as a point object.

(b) A monkey sitting on top of a man cycling smoothly on a circular track — Yes

If only the motion of the monkey around the circular track is considered, its size is negligible compared to the radius of the track, so it may be treated as a point object.

(c) A spinning cricket ball that turns sharply on hitting the ground — No

The spinning (rotational motion) of the ball affects its motion after bouncing, so its size and rotation cannot be ignored.

(d) A tumbling beaker that has slipped off the edge of a table — No

Since the beaker is tumbling, its rotational motion is significant, and it cannot be treated as a point object.

Answer: (a) and (b).

(a) A lives closer to the school than B.

(b) A starts from the school earlier than B.

(c) B walks faster than A.

(d) A and B reach home at the same time.

(e) B overtakes A on the road once.

Reason —

(a) The distance of A's home from the school is OP and that of B's home is OQ. Since OP is less than OQ, A lives closer to the school than B.

(b) For A, the position is x = 0 at time t = 0, but for B, x = 0 at a later time t > 0. Hence A starts from the school earlier than B.

(c) The speed is given by the slope of the position-time graph. Since the slope of the graph for B is greater than that for A, B walks faster than A.

(d) Corresponding to both the points P and Q, the value of time t is the same. Hence A and B both reach home at the same time.

(e) The graphs for A and B intersect each other only once, so A and B meet each other only once on the way home. Since B starts from the school later than A but walks faster than A, B overtakes A on the road once.

Given,

• Distance of the office from home, s = 2.5 km
• Walking speed while going, v₁ = 5 km h^-1
• Speed of the auto while returning, v₂ = 25 km h^-1
• Departure time from home = 9.00 am

To draw the graph we first work out the two travel times.

While going to the office, the time is the distance divided by the walking speed :

$$\begin{aligned}\text t_1 = \frac{\text s}{\text v_1} \\[1em] = \frac{2.5\ \text{km}}{5\ \text{km h}^{-1}} \\[1em] = \frac{1}{2}\ \text h = 30 \text { min}\end{aligned}$$

So she is at the office by 9.30 am. From 9.30 am until 5.00 pm she does not move, so her position stays fixed at 2.5 km and the graph runs flat during this long interval.

While coming back by auto, the return time is :

$$\begin{aligned}\text t_2 = \frac{\text s}{\text v_2} \\[1em] = \frac{2.5\ \text{km}}{25\ \text{km h}^{-1}} \\[1em] = \frac{1}{10}\ \text h = 6 \text { min}\end{aligned}$$

Hence she is back home by 5.06 pm.

Plotting position against time, the graph rises from 0 to 2.5 km between 9.00 and 9.30 am, stays constant at 2.5 km up to 5.00 pm, and then drops back to 0 by 5.06 pm, as shown.

The drunkard takes 37 s to fall into the pit.

Reason —

Given,

• Length of each step = 1 m
• Time for each step = 1 s
• Distance of the pit from the start = 13 m

In one full cycle the man first takes 5 steps ahead, advancing 5 m in 5 s, and then 3 steps back, losing 3 m in 3 s. So every cycle of 8 s carries him a net 2 m forward.

We must be careful near the end : the man falls in the moment he first reaches 13 m, even if that happens midway through a cycle.

Net gain per cycle = 5 - 3 = 2 m in 8 s.

Let us see how far he is after several complete cycles. To leave himself within one forward-swing of the pit, he needs to reach 8 m, because from 8 m a single set of 5 forward steps would carry him to 13 m.

Number of 8 s cycles needed to reach 8 m of net displacement :

$\text {cycles} = \frac{8\ \text m}{2\ \text m} = 4$

Time for these 4 cycles = 4 × 8 = 32 s, at the end of which he stands at 8 m.

From 8 m he now walks 5 steps forward without turning back. The remaining distance is 13 - 8 = 5 m, which takes 5 more seconds.

$\text {Total time} = 32\ \text s + 5\ \text s = 37 \text { s}$

The same value is read off the x-t graph, where the zig-zag line first touches the 13 m mark at t = 37 s.

Given,

• Initial speed, u = 126 km h^-1
• Final speed, v = 0 (the car stops)
• Stopping distance, s = 200 m

Before using the equations of motion we convert the speed into SI units :

$\text u = 126 \times \frac{1000\ \text m}{3600\ \text s} = 35 \text { m s}^{-1}$

Finding the retardation. The equation that links the two speeds with the distance (and leaves out time) is v² = u² + 2as. Rearranging for a :

$$\begin{aligned}\text a = \frac{\text v^2 - \text u^2}{2\text s} \\[1em] = \frac{(0)^2 - (35)^2}{2 \times 200} \\[1em] = \frac{-1225}{400} \\[1em] = -3.06 \text { m s}^{-2}\end{aligned}$$

The minus sign confirms that the acceleration opposes the motion, so it is a retardation of magnitude 3.06 m s^-2.

Finding the stopping time. Using v = u + at and solving for t :

$$\begin{aligned}\text t = \frac{\text v - \text u}{\text a} \\[1em] = \frac{0 - 35}{-3.06} \\[1em] = 11.43 \text { s}\end{aligned}$$

Hence the car decelerates at 3.06 m s^-2 and takes about 11.43 s to come to rest.

Given,

• Initial velocity, u = 29.4 m s^-1 (directed upward)
• Acceleration due to gravity, g = 9.8 m s^-2 (directed downward)

(a) Once the ball leaves the hand, the only influence on it is gravity. Gravity pulls everything toward the earth, so throughout the flight — while rising, at the top, and while falling — the acceleration points vertically downward.

(b) At the very top of its path the ball halts for an instant before turning round, so its speed there is zero. Gravity, however, never switches off. Therefore, at the highest point,

Velocity, v = 0 Acceleration, a = 9.8 m s^-2, directed vertically downward.

(c) Taking the upward direction as positive, at the highest point v = 0.

Height reached. Using v² = u² - 2gh with v = 0 :

$$\begin{aligned}0 = \text u^2 - 2\text {gh} \\[1em] \Rightarrow \text h = \frac{\text u^2}{2\text g} \\[1em] = \frac{(29.4)^2}{2 \times 9.8} \\[1em] = \frac{864.36}{19.6} \\[1em] = 44.1 \text { m}\end{aligned}$$

Time to reach the top. Using v = u - gt with v = 0 :

$$\begin{aligned}\text t = \frac{\text u}{\text g} \\[1em] = \frac{29.4}{9.8} \\[1em] = 3 \text { s}\end{aligned}$$

Since the downward journey mirrors the upward one, the ball needs the same 3 s to come down. Hence the total time before it lands back in the hand is

$\text T = 2\text t = 2 \times 3 = 6 \text { s}$

So the ball rises to 44.1 m and returns to the player's hands after 6 s.

(a) True. Zero speed at an instant only means that the velocity is zero at that instant; it does not tell us the rate of change of velocity. A ball thrown vertically upward has zero velocity at the highest point, but the acceleration due to gravity of 9.8 m s^-2 still acts on it at that instant.

(b) False. Speed is the magnitude of velocity. If the magnitude of the velocity is zero, the velocity itself is zero, since a velocity of zero magnitude cannot have any direction.

(c) True. In one-dimensional motion, a constant speed means that the magnitude and direction of the velocity both remain unchanged, so the acceleration is zero. (A change of direction while keeping the same speed would require an instantaneous reversal, that is, an infinite acceleration, which is not physically possible.)

(d) False. A positive acceleration increases the speed only when the body moves in the positive direction. If the body moves in the negative direction, a positive acceleration opposes the motion and decreases the speed. For example, a body moving with negative velocity and positive acceleration slows down.

Given,

• Height of fall, h = 90 m
• Initial velocity, u = 0
• Acceleration due to gravity, g = 9.8 m s^-2
• Fractional loss of speed at each bounce = one-tenth

Speed and time for the first fall. Using h = ut + $\frac{1}{2}$gt² with u = 0 :

$$\begin{aligned}\text h = 0+\frac{1}{2}\text {gt}^2 \\[1em] \Rightarrow \text t = \sqrt{\frac{2\text h}{\text g}} \\[1em] = \sqrt{\frac{2 \times 90}{9.8}} \\[1em] = \frac{30}{7} = 4.28 \text { s}\end{aligned}$$

The speed on striking the floor, from v = u + gt :

$\text v = 0 + 9.8 \times \frac{30}{7} = 42 \text { m s}^{-1}$

After the first bounce. The ball keeps nine-tenths of its speed :

$\text v' = \frac{9}{10} \times 42 = 37.8 \text { m s}^{-1}$

It rises with this speed and stops after a time t', where, from v = u - gt with final velocity 0 :

$\text t' = \frac{37.8}{9.8} = 3.86 \text { s}$

Coming down again it regains 37.8 m s^-1 at the floor and takes another 3.86 s. So the running total of time is

$\text t + \text t' + \text t'' = 4.28 + 3.86 + 3.86 = 12 \text { s}$

Just after the second bounce the speed becomes

$\text v'' = \frac{9}{10} \times 37.8 = 34.0 \text { m s}^{-1}$

Collecting these values :

Time	Speed
0	0
4.28 s	42 m s-1 → 37.8 m s-1
8.14 s	0
12 s	37.8 m s-1 → 34 m s-1

Plotting speed against time gives a series of straight sloping lines: the speed climbs at 9.8 m s^-1 per second during each fall and drops at the same rate during each rise, with a sudden step down at every bounce. The graph is drawn as shown.

(a) Magnitude of displacement and total length of path.

The magnitude of displacement is the shortest distance between the initial and final positions of the particle, and displacement is a vector quantity. The total length of path is the actual distance covered by the particle along its path, and it is a scalar quantity.

Example : Let a particle move from A (0 m) to B (4 m) and then return to C (2 m).

• Magnitude of displacement = final position - initial position = 2 - 0 = 2 m.
• Total length of path = (A to B) + (B to C) = 4 + 2 = 6 m.

Since the total length of path includes the whole distance covered while the magnitude of displacement is only the net change in position, the total length of path is always greater than or equal to the magnitude of displacement.

The two are equal when the particle moves along a straight line in one direction without reversing. If it moves straight from A (0 m) to B (4 m), both quantities are equal to 4 m.

(b) Magnitude of average velocity and average speed.

$\text {Average velocity} = \frac{\text {displacement}}{\text {time interval}}, \qquad \text {Average speed} = \frac{\text {total length of path}}{\text {time interval}}$

Example : Let the same particle take 2 s from A (0 m) to B (4 m) and 1 s from B to C (2 m).

Displacement = 2 m, total length of path = 6 m, total time interval = 3 s.

$\text {Magnitude of average velocity} = \frac{2\ \text m}{3\ \text s} = 0.67 \text { m s}^{-1}$

$\text {Average speed} = \frac{6\ \text m}{3\ \text s} = 2 \text { m s}^{-1}$

Since both are divided by the same time interval, and the total length of path is always greater than or equal to the magnitude of displacement, the average speed is always greater than or equal to the magnitude of average velocity.

The two are equal only for motion along a straight line in one direction without reversal. For example, moving from A to B (4 m in 2 s) gives average velocity = average speed = 2 m s^-1.

Given,

• One-way distance, home to market, s = 2.5 km
• Speed while going, v₁ = 5 km h^-1
• Speed while returning, v₂ = 7.5 km h^-1

First we find how long each leg of the trip lasts.

Time to reach the market :

$\text t_1 = \frac{\text s}{\text v_1} = \frac{2.5\ \text{km}}{5\ \text{km h}^{-1}} = 0.5 \text { h} = 30 \text { min}$

Time to walk back :

$\text t_2 = \frac{\text s}{\text v_2} = \frac{2.5\ \text{km}}{7.5\ \text{km h}^{-1}} = \frac{1}{3} \text { h} = 20 \text { min}$

So he reaches the market at the 30 min mark and is home again at the 50 min mark.

(i) From 0 to 30 min. He is still on his way out, having covered 2.5 km in one direction. Here distance and displacement are both 2.5 km.

$\text {Magnitude of average velocity} = \frac{2.5\ \text{km}}{0.5\ \text h} = 5 \text { km h}^{-1}$

$\text {Average speed} = \frac{2.5\ \text{km}}{0.5\ \text h} = 5 \text { km h}^{-1}$

(ii) From 0 to 50 min. By now he has gone out and come back, ending where he started.

Displacement = 0, and distance = 2.5 + 2.5 = 5 km.

Time = 50 min = $\frac{5}{6}$ h.

$\text {Magnitude of average velocity} = \frac{0}{(5/6)\ \text h} = 0$

$\text {Average speed} = \frac{5\ \text{km}}{(5/6)\ \text h} = 6 \text { km h}^{-1}$

(iii) From 0 to 40 min. The first 30 min take him 2.5 km out; the next 10 min bring him partway back. In those 10 min at 7.5 km h^-1 he covers

$7.5 \times \frac{10}{60} = 1.25 \text { km}$

So displacement = 2.5 - 1.25 = 1.25 km, and distance = 2.5 + 1.25 = 3.75 km.

Time = 40 min = $\frac{2}{3}$ h.

$\text {Magnitude of average velocity} = \frac{1.25\ \text{km}}{(2/3)\ \text h} = 1.875 \text { km h}^{-1}$

$\text {Average speed} = \frac{3.75\ \text{km}}{(2/3)\ \text h} = 5.625 \text { km h}^{-1}$

The distinction between average speed and magnitude of average velocity arises because, over a time interval, the total length of path and the magnitude of displacement can be different. The instantaneous speed and the magnitude of instantaneous velocity are measured over an infinitesimally small time interval, and in such a small interval the particle cannot reverse its direction. Over this interval the length of path covered and the magnitude of displacement become equal, so the instantaneous speed is always equal to the magnitude of instantaneous velocity.

Example : If a car has an instantaneous velocity of + 10 m s^-1 or - 10 m s^-1 at an instant, its instantaneous speed at that instant is 10 m s^-1 in both cases; only the direction (sign) is left out.

None of the four graphs can describe the one-dimensional motion of a particle.

(a) A vertical line drawn at some instant meets the curve at two places, implying the particle occupies two positions at the same time. A single particle cannot be in two spots at once, so this is impossible. (The arrows on the curve carry no physical meaning.)

(b) Here too a vertical line cuts the graph twice, so the particle would need a positive and a negative velocity simultaneously — motion in two opposite directions at the same instant. That cannot happen in one-dimensional motion.

(c) The curve dips below the axis, showing negative speeds at certain moments. Speed is the magnitude of velocity and can never be negative, so this graph is unphysical.

(d) After a while the total path length L on this graph begins to fall. Path length is a running total of distance covered and can only increase or stay constant; it can never decrease. Hence this graph is impossible.

No, it is not correct.

A position-time (x-t) graph shows how the position of the particle changes with time; it does not represent the actual path of the particle in space. Hence the shape of the graph, whether straight or parabolic, describes the variation of position with time, not the geometrical path followed by the particle.

A suitable physical context for this graph is a particle at rest at x = 0 up to t = 0, which is then released and falls freely under gravity. During the free fall the distance covered is directly proportional to the square of time, which gives the parabolic part of the graph for t > 0, while for t < 0 the particle remains at rest at the origin.

Given,

• Speed of the police van, v_van = 30 km h^-1
• Speed of the thief's car, v_thief = 192 km h^-1
• Muzzle speed of the bullet (relative to the van), v_muzzle = 150 m s^-1

First convert the two vehicle speeds into SI units :

$\text v_\text{van} = 30 \times \frac{1000}{3600} = 8.33 \text { m s}^{-1}$

$\text v_\text{thief} = 192 \times \frac{1000}{3600} = 53.33 \text { m s}^{-1}$

Step 1 — Speed of the bullet relative to the ground. The bullet leaves the gun at 150 m s^-1 relative to the van, and the van is itself moving forward, so the two speeds add :

$\text v_\text{bullet, ground} = \text v_\text{van} + \text v_\text{muzzle} = 8.33 + 150 = 158.33 \text { m s}^{-1}$

Step 2 — Speed of the bullet relative to the thief's car. The damage-causing speed is the bullet's speed as seen from the fleeing car, which is the difference of the two ground speeds (both point the same way) :

$\text v_\text{bullet, thief} = \text v_\text{bullet, ground} - \text v_\text{thief} = 158.33 - 53.33 = 105 \text { m s}^{-1}$

Hence the bullet strikes the thief's car at a relative speed of 105 m s^-1.

(a) In this position-time graph, the position x remains zero for some time, then increases to a maximum value A, then decreases at a smaller rate to zero, then increases in the opposite direction (negative) at the same smaller rate, and finally becomes constant at a value B. This may represent a ball lying at rest on a smooth floor which is kicked towards a wall; it rebounds from the wall with reduced speed, moves to the opposite wall and comes to rest there.

(b) In this velocity graph, the velocity changes its direction repeatedly, and its magnitude decreases each time. This may represent a ball thrown vertically upward which rebounds repeatedly from the floor with reduced velocity after each impact.

(c) In this acceleration graph, the acceleration is zero except for a short interval of time. This may represent a ball moving with uniform velocity which is struck by a bat for a very short interval of time, the short-lived acceleration being the impact that reverses its motion.

For a particle in simple harmonic motion, the acceleration always points back toward the mean position, that is, opposite to the displacement x. Also, the slope of the x-t graph at any moment gives the velocity there. Using these two ideas :

At t = 0.3 s : The graph shows x below the axis, so x is negative (x < 0). The slope at this point is negative, so v is negative (v < 0). Since acceleration opposes displacement, a is positive (a > 0).

At t = 1.2 s : Here x is above the axis, so x is positive (x > 0), and the slope is positive, so v is positive (v > 0). Acceleration, being opposite to x, is negative (a < 0).

At t = - 1.2 s : The curve gives x negative (x < 0) with a positive slope, so v is positive (v > 0). Since x is negative, the acceleration is positive (a > 0).

On a position-time graph, the steepness (magnitude of the slope) over a small stretch tells us the average speed there. Reading the three marked stretches, the curve is steepest in interval 3 and flattest in interval 2. Therefore the average speed is greatest in interval 3 and least in interval 2.

For the direction, we look at the sign of the slope. The graph rises (positive slope) through intervals 1 and 2, but falls (negative slope) in interval 3. So the average velocity is positive in intervals 1 and 2 and negative in interval 3. Speed stays positive throughout, since it ignores direction.

Greatest average acceleration. Average acceleration over an interval is the change in speed divided by the time, $\frac{\Delta \text v}{\Delta \text t}$. As the three intervals last equally long, we just compare how steeply the speed curve rises or falls in each.

• Interval 1 (A to B) : speed rises → positive acceleration.
• Interval 2 (B to C) : speed drops sharply → negative acceleration, and this fall is the steepest.
• Interval 3 (C to D) : speed rises again → positive acceleration, but less steeply than in interval 1.

Because interval 2 has the steepest change, the average acceleration is greatest in magnitude in interval 2 (B to C).

Greatest average speed. With direction fixed, the average speed over an interval goes with the area under the speed-time curve for that interval.

• Interval 1 : a moderate area.
• Interval 2 : the speed is falling, so the area is small.
• Interval 3 : high speeds are held for the whole interval, giving the largest area.

So the average speed is greatest in interval 3 (C to D).

Signs of v and a. The motion never reverses, so the speed is positive throughout (v > 0); only a changes sign.

S. No.	Interval	Sign of speed (v)	Sign of acceleration (a)
1.	(A - B)	+	+ (speed rising)
2.	(B - C)	+	- (speed falling)
3.	(C - D)	+	+ (speed rising)

Accelerations at A, B, C, D. The acceleration at a point is the slope of the speed-time curve there.

• Point A : the curve is just starting to rise → positive slope ⇒ positive acceleration.
• Point B : a crest (maximum speed) ⇒ zero slope ⇒ acceleration = 0.
• Point C : a trough (minimum speed) ⇒ zero slope ⇒ acceleration = 0.
• Point D : another crest ⇒ zero slope ⇒ acceleration = 0.

Thus the acceleration is positive at A and zero at each of B, C and D.

Given,

• Speed of the jet plane relative to the ground, magnitude = 500 km h^-1
• Speed of the combustion products relative to the plane, v_CJ = 1500 km h^-1

Let us fix a sign convention : take the direction pointing toward the ground observer as positive, and assume the jet is flying away from that observer.

With this choice, the jet's ground velocity points away from the observer :

$\vec{\text v}_\text J = -500 \text { km h}^{-1}$

The exhaust is thrown out backwards, that is, toward the observer, so relative to the plane

$\vec{\text v}_\text{CJ} = +1500 \text { km h}^{-1}$

The velocity of the combustion products relative to the plane is, by definition,

$$\begin{aligned}\vec{\text v}_\text{CJ} = \vec{\text v}_\text C - \vec{\text v}_\text J \\[1em] \Rightarrow \vec{\text v}_\text C = \vec{\text v}_\text{CJ} + \vec{\text v}_\text J\end{aligned}$$

Substituting the values,

$\vec{\text v}_\text C = 1500 + (-500) = 1000 \text { km h}^{-1}$

Hence the combustion products move at 1000 km h^-1 with respect to the ground observer.

Given,

• Length of each train = 400 m, so l_A = l_B = 400 m
• Initial speed of both trains, u_A = u_B = 72 km h^-1 = 20 m s^-1
• Acceleration of train B, a_B = 1 m s^-2
• Acceleration of train A, a_A = 0
• Time, t = 50 s

Let d be the original distance between the front of train B and the rear of train A.

Since both trains have the same initial speed, the initial relative velocity of B with respect to A is

$\text u_\text{rel} = \text u_\text B - \text u_\text A = 20 - 20 = 0$

and the relative acceleration of B with respect to A is

$\text a_\text{rel} = \text a_\text B - \text a_\text A = 1 - 0 = 1 \text { m s}^{-2}$

The distance travelled by train B relative to train A is obtained by using the equation s = ut + $\frac{1}{2}$at² for the relative motion,

$$\begin{aligned}\text s = \text u_\text{rel}\text t + \frac{1}{2}\text a_\text{rel}\text t^2 \\[1em] = (0 \times 50) + \frac{1}{2} \times 1 \times (50)^2 \\[1em] = \frac{1}{2} \times 2500 \\[1em] = 1250 \text { m}\end{aligned}$$

So train B gains 1250 m on train A in 50 s.

For the guard (rear end) of train B to just move past the driver (front end) of train A, this relative distance must be equal to the original distance between the trains plus the length of both the trains :

$$\begin{aligned}\text s = \text d + \text l_\text A + \text l_\text B \\[1em] 1250 = \text d + 400 + 400 \\[1em] 1250 = \text d + 800 \\[1em] \text d = 450 \text { m}\end{aligned}$$

Hence, the original distance between the two trains was 450 m.

Given,

• Cyclist's speed, v_C = 20 km h^-1 (from A toward B)
• A bus overtakes him from behind every 18 min
• A bus meets him head-on every 6 min
• Buses leave each terminus every T min

Let v_B be the constant speed of the buses. The spacing between two consecutive buses on the road is the distance one bus covers in T minutes, that is,

$\text {spacing} = \text v_\text B \times \frac{\text T}{60} \text { km}$

Buses coming from behind (same direction as the cyclist). A following bus approaches the cyclist at the relative speed (v_B - v_C), and it takes 18 min to complete the spacing between two buses :

$$\begin{aligned}(\text v_\text B - \text v_\text C) \times \frac{18}{60} = \text v_\text B \times \frac{\text T}{60} \\[1em] \Rightarrow \text v_\text B - \text v_\text C = \frac{\text v_\text B \text T}{18} \quad \text{...(i)}\end{aligned}$$

Buses coming head-on (opposite direction). An oncoming bus approaches at the relative speed (v_B + v_C), taking only 6 min to cover the same spacing :

$$\begin{aligned}(\text v_\text B + \text v_\text C) \times \frac{6}{60} = \text v_\text B \times \frac{\text T}{60} \\[1em] \Rightarrow \text v_\text B + \text v_\text C = \frac{\text v_\text B \text T}{6} \quad \text{...(ii)}\end{aligned}$$

Solving for v_B. Dividing (ii) by (i) cancels the unknown T :

$$\begin{aligned}\frac{\text v_\text B + \text v_\text C}{\text v_\text B - \text v_\text C} = \frac{\text v_\text B \text T / 6}{\text v_\text B \text T / 18} = \frac{18}{6} = 3 \\[1em] \Rightarrow \text v_\text B + \text v_\text C = 3(\text v_\text B - \text v_\text C) \\[1em] \Rightarrow \text v_\text B + \text v_\text C = 3\text v_\text B - 3\text v_\text C \\[1em] \Rightarrow 2\text v_\text B = 4\text v_\text C \\[1em] \Rightarrow \text v_\text B = 2\text v_\text C = 2 \times 20 = 40 \text { km h}^{-1}\end{aligned}$$

Solving for T. Putting v_B = 40 and v_C = 20 into (ii) :

$$\begin{aligned}40 + 20 = \frac{40 \times \text T}{6} \\[1em] 60 = \frac{40\text T}{6} \\[1em] \text T = \frac{60 \times 6}{40} = 9 \text { min}\end{aligned}$$

Hence the buses run at 40 km h^-1 and one leaves each terminus every 9 min.

Given,

• Maximum launch speed of the ball, u = 49 m s^-1 (upward)
• Uniform speed of the lift in the second case, v_lift = 5 m s^-1 (upward)
• Acceleration due to gravity, g = 9.8 m s^-2

Case 1 — the lift is stationary. Let t be the time to reach the top of the flight, where the ball's velocity is zero. Taking upward as positive, from v = u - gt with v = 0 :

$$\begin{aligned}0 = \text u - \text {gt} \\[1em] \Rightarrow \text t = \frac{\text u}{\text g} = \frac{49}{9.8} = 5 \text { s}\end{aligned}$$

The descent takes just as long, so the total time back to the hand is

$\text T = 2\text t = 2 \times 5 = 10 \text { s}$

Case 2 — the lift rises at a steady 5 m s^-1. What matters for catching the ball is the motion of the ball relative to the boy. Because the lift moves at a constant velocity (no acceleration of its own), the boy's frame is still an inertial one. In that frame the ball again leaves at 49 m s^-1 and feels the same gravity, so its up-and-down journey relative to the boy is unchanged.

$\therefore\ \text T = 10 \text { s}$

Hence the ball comes back to the boy's hands after 10 s in both cases.

Given,

• Speed of the child relative to the belt, v_CB = 9 km h^-1
• Speed of the belt relative to the ground, v_B = 4 km h^-1
• Separation between the parents on the belt = 50 m

Take the belt's direction of travel (left to right) as positive.

(a) Child running the same way as the belt. For a ground observer, the child's velocity is the sum of the child's velocity relative to the belt and the belt's own velocity :

$\vec{\text v}_\text C = \vec{\text v}_\text{CB} + \vec{\text v}_\text B = 9 + 4 = 13 \text { km h}^{-1}$

(b) Child running against the belt. Now the child's velocity relative to the belt is negative, v_CB = - 9 km h^-1 :

$\vec{\text v}_\text C = \vec{\text v}_\text{CB} + \vec{\text v}_\text B = -9 + 4 = -5 \text { km h}^{-1}$

The negative sign means the child is seen moving right to left.

(c) Time for each run. The parents stand on the same belt as the child, so relative to them the child always moves at 9 km h^-1 whichever way he runs. The 50 m separation is a belt distance, so we use the child's belt speed :

$\text t = \frac{50\ \text m}{9\ \text{km h}^{-1}} = \frac{0.05\ \text{km}}{9\ \text{km h}^{-1}} = \frac{0.05}{9} \times 3600 = 20 \text { s}$

So each crossing takes 20 s in either direction.

Effect of viewing from a parent. A parent shares the belt's motion, so the belt's speed no longer enters. From a parent's viewpoint the child moves at 9 km h^-1 both ways, so the answers to (a) and (b) change (both become 9 km h^-1), while the time found in (c) stays the same.

Given,

• Height of the cliff, x₀ = 200 m
• Initial speed of stone 1, u₁ = 15 m s^-1 (upward)
• Initial speed of stone 2, u₂ = 30 m s^-1 (upward)
• Acceleration due to gravity, g = 10 m s^-2

Take upward as positive and measure heights from the ground.

When does stone 1 land? Its height at time t is

$\text x_1 = \text x_0 + \text u_1\text t - \frac{1}{2}\text {gt}^2 = 200 + 15\text t - 5\text t^2 \quad \text{...(i)}$

Setting x₁ = 0 for the ground :

$$\begin{aligned}5\text t^2 - 15\text t - 200 = 0 \\[1em] \text t^2 - 3\text t - 40 = 0 \\[1em] (\text t - 8)(\text t + 5) = 0 \\[1em] \Rightarrow \text t = 8 \text { s} \\ (\text{rejecting } \text t = -5 \text { s})\end{aligned}$$

So stone 1 reaches the ground at t = 8 s.

When does stone 2 land? Its height at time t is

$\text x_2 = \text x_0 + \text u_2\text t - \frac{1}{2}\text {gt}^2 = 200 + 30\text t - 5\text t^2 \quad \text{...(ii)}$

Setting x₂ = 0 :

$$\begin{aligned}5\text t^2 - 30\text t - 200 = 0 \\[1em] \text t^2 - 6\text t - 40 = 0 \\[1em] (\text t - 10)(\text t + 4) = 0 \\[1em] \Rightarrow \text t = 10 \text { s} \\ (\text{rejecting } \text t = -4 \text { s})\end{aligned}$$

So stone 2 reaches the ground at t = 10 s.

Relative position, 0 to 8 s. While both stones are airborne, subtract (i) from (ii). The gravity terms cancel :

$\text x_2 - \text x_1 = (\text u_2 - \text u_1)\text t = (30 - 15)\text t = 15\text t$

This is a straight line, so from t = 0 to t = 8 s the separation grows linearly — the segment OA of the graph. At t = 8 s it reaches 15 × 8 = 120 m.

Relative position, 8 s to 10 s. After 8 s, stone 1 is lying on the ground (x₁ = 0) while stone 2 is still falling. The separation is now just x₂ itself :

$\text x_2 - \text x_1 = 200 + 30\text t - 5\text t^2$

Being quadratic, this is the curved part AB of the graph, and it falls back to zero when stone 2 lands at t = 10 s.

After 10 s. Both stones rest on the ground, so their separation stays zero — the flat part BC.

The described straight line (0 to 8 s), curve (8 to 10 s) and flat line (beyond 10 s) match the given graph, verifying it. The equations are x₂ - x₁ = 15 t for the linear part and x₂ - x₁ = 200 + 30 t - 5 t² for the curved part.

Given (read from the speed-time graph),

• The speed rises from 0 at t = 0 to a peak of 12 m s^-1 at t = 5 s, then falls back to 0 at t = 10 s.

(a) Distance from t = 0 to t = 10 s. On a speed-time graph the distance equals the area beneath the curve, here the triangle OAB with base OB = 10 s and height AC = 12 m s^-1 :

$$\begin{aligned}\text {Distance} = \frac{1}{2} \times \text {base} \times \text {height} \\[1em] = \frac{1}{2} \times 10 \times 12 \\[1em] = 60 \text { m}\end{aligned}$$

Average speed over this interval :

$\frac{\text {distance}}{\text {time}} = \frac{60\ \text m}{10\ \text s} = 6 \text { m s}^{-1}$

(b) Distance from t = 2 s to t = 6 s. This interval straddles the peak, so we split it at t = 5 s into a rising part (2 s to 5 s) and a falling part (5 s to 6 s).

Rising part. The acceleration on OA is the slope :

$\text a_1 = \frac{12 - 0}{5 - 0} = 2.4 \text { m s}^{-2}$

Speed at t = 2 s :

$\text u_1 = 0 + \text a_1 \times 2 = 2.4 \times 2 = 4.8 \text { m s}^{-1}$

Distance from t = 2 s to t = 5 s (duration 3 s) :

$$\begin{aligned}\text s_1 = \text u_1\text t + \frac{1}{2}\text a_1\text t^2 \\[1em] = 4.8 \times 3 + \frac{1}{2} \times 2.4 \times (3)^2 \\[1em] = 14.4 + 10.8 = 25.2 \text { m}\end{aligned}$$

Falling part. The acceleration on AB is

$\text a_2 = \frac{0 - 12}{10 - 5} = -2.4 \text { m s}^{-2}$

Starting from the peak speed 12 m s^-1, distance from t = 5 s to t = 6 s (duration 1 s) :

$$\begin{aligned}\text s_2 = \text u_2\text t + \frac{1}{2}\text a_2\text t^2 \\[1em] = 12 \times 1 + \frac{1}{2} \times (-2.4) \times (1)^2 \\[1em] = 12 - 1.2 = 10.8 \text { m}\end{aligned}$$

Total distance from t = 2 s to t = 6 s :

$\text s = \text s_1 + \text s_2 = 25.2 + 10.8 = 36 \text { m}$

Average speed over this interval :

$\frac{36\ \text m}{(6 - 2)\ \text s} = 9 \text { m s}^{-1}$

Formulae 3, 4 and 6 are correct; formulae 1, 2 and 5 are wrong.

Reason —

The v-t graph shown is curved, which means the acceleration is not constant between t₁ and t₂.

1. Wrong. The term $\frac{1}{2}$a(t₂ - t₁)² comes from the equation of motion for uniform acceleration. As the acceleration varies here, this relation does not apply.

2. Wrong. v = u + a(t₂ - t₁) again assumes a single fixed acceleration, which is not the case for this curved graph.

3. Correct. This is simply the definition of average velocity — total displacement over total time — and holds no matter how the acceleration behaves.

4. Correct. This is the definition of average acceleration — total change in velocity over total time — and is always valid.

5. Wrong. Slotting the average acceleration into a uniform-acceleration formula is not legitimate, so this expression does not describe the motion.

6. Correct. The area enclosed by the v-t curve and the time-axis always equals the displacement, whether or not the acceleration is uniform.

Given,

• Height of the aircraft above the observer, OC = 3400 m
• Angle subtended at the observer by the two positions = 30°
• Time between the two positions = 10 s

Let O be the observation point on the ground and let the aircraft fly horizontally along PQ, passing positions A and B that are 10 s apart. C is the foot of the perpendicular from O to the flight line, so OC = 3400 m.

By symmetry the line OC bisects the 30° angle, giving

$\angle \text{AOC} = \angle \text{COB} = 15°$

and the aircraft covers AC in half of 10 s, that is, 5 s.

Finding AC. In the right-angled triangle ACO,

$$\begin{aligned}\tan 15° = \frac{\text {AC}}{\text {OC}} \\[1em] \Rightarrow \text {AC} = \text {OC} \times \tan 15° \\[1em] = 3400 \times 0.2679 \\[1em] = 911 \text { m}\end{aligned}$$

Finding the speed. The aircraft travels this 911 m in 5 s, so

$\text v = \frac{\text {AC}}{\text {time}} = \frac{911\ \text m}{5\ \text s} = 182.2 \text { m s}^{-1}$

Hence the aircraft is flying at about 182.2 m s^-1.

A reference frame can be stationary or in motion.

Reason — A frame of reference is a chosen set of axes together with a clock, used as a viewpoint for describing motion. Nothing forces this viewpoint to be at rest, so it may be either stationary or moving.

The velocity of objects remains constant unless acted upon by a force.

Reason — An inertial frame is one in which Newton's first law holds. That law states a body keeps its state of rest or of uniform motion until some external force changes it — so velocity stays constant unless a force acts.

Instantaneous speed

Reason — The steepness of the distance-time graph at a single point measures how fast distance is changing right there, which is exactly the instantaneous speed.

Displacement

Reason — Multiplying velocity by time gives displacement, and that product is precisely the area trapped between the velocity-time line and the time-axis.

Zero

Reason — The slope of a velocity-time graph equals the acceleration. With uniform velocity there is no change in velocity, so the acceleration and hence the slope are both zero.

A straight line passing through the origin

Reason — Uniform acceleration means the velocity grows by equal amounts in equal times, tracing a straight line; starting from rest it passes through the origin.

v = u + at

Reason — The first equation of motion relates the final velocity (v), initial velocity (u), acceleration (a), and time (t):

v = u + at

The other options represent different equations of motion:

• s = ut + $\frac{1}{2}$at² — Second equation of motion
• v² = u² + 2as — Third equation of motion
• v = at — Valid only when the initial velocity u = 0, so it is not the general first equation of motion.

v² = u² + 2as

Reason — The third equation of motion links the two velocities with the acceleration and the displacement, without any reference to time, and is written v² = u² + 2as.

F = ma

Reason — The other three relations tie together displacement, velocity, acceleration and time, so they are kinematic. F = ma expresses Newton's second law and belongs to dynamics, not kinematics.

0 m/s

Reason — At the top of its rise the stone reverses direction, and for that brief instant it is neither going up nor coming down, so its velocity there is zero.

The acceleration is constant and equals g.

Reason — A freely falling body is acted on by gravity alone, which imparts a steady downward acceleration equal to g throughout the fall.

Speed

Reason — In uniform circular motion the body moves at a fixed speed, but its direction — and therefore its velocity and acceleration — keeps changing. Only the speed stays constant.

Zero

Reason — A particle moving in a straight line with a constant speed has no change in the magnitude or direction of its velocity. Since the velocity remains constant, its acceleration is zero.

It is the same as average velocity over a time interval.

Reason — Instantaneous velocity is the velocity at a particular instant, whereas average velocity is measured over a time interval. The two are generally not equal, so this is not a characteristic of instantaneous velocity.

Displacement

Reason — The region enclosed between the velocity-time graph and the time-axis represents the displacement covered by the object.

It can be zero even if the distance is non-zero

Reason — Displacement depends only on the start and end points. If a body returns to where it began, its displacement is zero even though it has travelled a non-zero distance.

50 km/h

Reason —

Given,

• Distance travelled, s = 150 km
• Time taken, t = 3 h

$\text {Average speed} = \frac{\text {Distance travelled}}{\text {Time taken}} = \frac{150\ \text {km}}{3\ \text h} = 50 \text { km h}^{-1}$

The particle is moving with uniform speed

Reason — If the instantaneous speed never differs from the average speed, the speed is not changing at all, which is the sign of uniform-speed motion.

It is the rate of change of velocity with respect to time at a specific instant.

Reason — Instantaneous acceleration is defined as how fast the velocity is changing at one particular moment in time.

5 m/s²

Reason —

Given,

• Initial velocity, u = 0 (starts from rest)
• Final velocity, v = 25 m s^-1
• Time, t = 5 s

From v = u + at, the acceleration is

$\text a = \frac{\text v - \text u}{\text t} = \frac{25 - 0}{5} = 5 \text { m s}^{-2}$

2 s

Reason —

Given,

• Height, h = 20 m
• Initial velocity, u = 0
• g = 10 m s^-2

From h = ut + $\frac{1}{2}$gt², the time of fall is

$$\begin{aligned}\text{h} = (0\times \text{t}) +\frac{1}{2}\text{gt}^2\\[1em] \text t = \sqrt{\frac{2\text h}{\text g}} \\[1em] \text t = \sqrt{\frac{2 \times 20}{10}} \\[1em] \text t = \sqrt{4} \\[1em] \text t = 2 \text { s}\end{aligned}$$

It covers equal distances in equal intervals of time.

Reason — Uniform velocity means the body advances the same distance in every equal interval of time, along an unchanging direction.

Uniformly accelerated

Reason — The particle covers 20 m in the first 2 seconds and 40 m in the next 2 seconds. Since the distance covered increases by equal amounts in equal intervals of time, the velocity increases uniformly, so the motion is uniformly accelerated.

12 km

Reason —

Given,

• Uniform velocity, v = 40 m s^-1
• Time, t = 5 min = 300 s

$\text {Distance} = \text v \times \text t = 40 \times 300 = 12000 \text { m} = 12 \text { km}$

Note: The textbook’s printed answer key gives option 1, which is 10 km. However, the book’s own solution hint calculates the distance as 40 m s⁻¹ × 300 s = 12,000 m = 12 km. Therefore, option 2 is correct, and option 1 in the printed answer key is a misprint.

t²

Reason — Starting from rest, the distance is s = $\text {ut}+\frac{1}{2}\text{at}^2\implies\frac{1}{2}\text{at}^2$. With the acceleration $a$ held constant, $s$ varies as the square of the time, so s ∝ t².

It halves

Reason — For a fixed distance, the time taken is inversely proportional to the speed. Hence, if the speed of the car doubles, the time taken to cover the same distance becomes half.

Constant and positive acceleration

Reason —

Given the displacement y = kt² with k a positive constant, differentiate once to get the velocity :

$\text v = \frac{\text {dy}}{\text {dt}} = 2\text {kt}$

Since v depends on t and k is positive, the velocity is positive and time-dependent (not constant). Differentiating again gives the acceleration :

$\text a = \frac{\text {dv}}{\text {dt}} = 2\text k$

This is a positive constant, so the body has a constant, positive acceleration.

5 m/s

Reason —

Given the position x = 5t + 6t², the velocity at any instant is

$\text v = \frac{\text {dx}}{\text {dt}} = 5 + 12\text t$

The initial velocity is the value at t = 0 :

$\text u = 5 + 12(0) = 5 \text { m s}^{-1}$

3 m/s²

Reason —

Given,

• Initial velocity, u = 20 m s^-1
• Final velocity, v = 50 m s^-1
• Time, t = 10 s

$\text {Average acceleration} = \frac{\text v - \text u}{\text t} = \frac{50 - 20}{10} = 3 \text { m s}^{-2}$

20 m/s

Reason —

Given,

• Initial velocity, u = 10 m s^-1
• Acceleration, a = 2 m s^-2
• Time, t = 5 s

From v = u + at,

$\text v = 10 + (2 \times 5) = 10 + 10 = 20 \text { m s}^{-1}$

Note: The textbook’s printed answer key gives option 4, which is 25 m s⁻¹. However, using the formula $v = u + at$, we get $v = 10 + (2 \times 5) = 20$ m s⁻¹. Therefore, option 1 is correct, and option 4 in the printed answer key is a misprint.

150 m

Reason —

Given,

• Initial velocity, u = 0 (starts from rest)
• Acceleration, a = 3 m s^-2
• Time, t = 10 s

From s = ut + $\frac{1}{2}$at²,

$$\begin{aligned}\text s = (0 \times 10) + \frac{1}{2} \times 3 \times (10)^2 \\[1em] = \frac{1}{2} \times 3 \times 100 \\[1em] = 150 \text { m}\end{aligned}$$

s = ut + $\frac{1}{2}$at²

Reason — For motion at constant acceleration, the displacement in time t is given by the second equation of motion, s = ut + $\frac{1}{2}$at².

Constant

Reason — The slope of a displacement-time graph is the velocity. When the velocity is uniform, that slope does not change, so it stays constant.

Its velocity changes at a uniform rate.

Reason — Constant acceleration means the velocity gains (or loses) the same amount every second, that is, it changes at a uniform rate.

Constant

Reason — Uniform acceleration gives a straight-line velocity-time graph, and a straight line has the same slope everywhere, so the slope is constant.

Both (2) and (3).

Reason — A horizontal line on a velocity-time graph shows that the velocity is constant. The slope of this line is zero, and since the slope of a velocity-time graph gives the acceleration, the acceleration is zero. Hence both constant velocity and zero acceleration are represented.

A rotating merry-go-round

Reason — A spinning merry-go-round is constantly changing the direction of motion of everything on it, so it is accelerating. A frame that accelerates is non-inertial.

Average speed is always greater than or equal to average velocity.

Reason — Average speed is the total distance divided by the total time, while average velocity is the total displacement divided by the total time. Since the distance is always greater than or equal to the magnitude of displacement, the average speed is always greater than or equal to the magnitude of average velocity.

Uniform

Reason — A displacement-time graph that is a straight line tilted to the time-axis has a fixed slope, which means a constant velocity, that is, uniform motion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When an object returns to its initial position, its displacement is zero, even though the distance travelled by it along the path is not zero.

Reason (R) is also correct: Displacement is a vector quantity that depends only on the initial and final positions of the object and not on the path taken, whereas distance is a scalar quantity that depends on the actual path covered. This is why the displacement can be zero while the distance is non-zero, so the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The velocity of an object can be zero at an instant while its acceleration is non-zero. For example, a ball thrown vertically upward has zero velocity at the highest point but still has a non-zero acceleration due to gravity.

Reason (R) is also correct: Acceleration is the rate of change of velocity with respect to time, and a changing velocity can become zero at a particular instant. Hence the velocity can be zero while the acceleration is non-zero, so the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Uniform motion keeps the velocity fixed, and a fixed velocity plots as a horizontal line — one that runs parallel to the time axis.

Reason (R) is also correct: Covering equal distances in equal times is precisely what holds the velocity constant, which produces that horizontal graph. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Two bodies going the same way at matching speeds keep a fixed separation, so neither sees the other move — their relative velocity is zero.

Reason (R) is also correct: Relative velocity is what one body's motion looks like from the other, obtained as the difference of their velocities; for equal, same-direction velocities that difference is zero. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: The area beneath a velocity-time graph gives the displacement, not the acceleration.

Reason (R) is also false: It is the slope of a velocity-time graph that gives the acceleration, not the displacement.

Therefore, both assertion and reason are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: From launch to landing the ball is governed by gravity alone, so its acceleration keeps the same value g the whole time.

Reason (R) is also correct: In free fall the only force is gravity, and it delivers a steady downward acceleration of about 9.8 m s^-2. This is why the acceleration stays constant, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In non-uniform motion the acceleration is not constant, so the equation s = ut + $\frac{1}{2}$at² cannot be used, and other methods such as calculus are required to calculate the displacement.

Reason (R) is also correct: The equation s = ut + $\frac{1}{2}$at² is derived under the assumption of uniform acceleration. This assumption is the reason it cannot be applied to non-uniform motion, so the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When an object moves with uniform velocity, its velocity remains constant in both magnitude and direction. Therefore, the instantaneous velocity at any instant is equal to the average velocity over any time interval.

Reason (R) is also correct: In uniform velocity, the object covers equal displacements in equal intervals of time (not just equal distances), and its direction of motion remains unchanged. Since the velocity remains constant throughout the motion, the average velocity over any interval is the same as the instantaneous velocity at every instant. Thus, the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When the velocity is non-uniform the slope of the distance-time graph keeps changing, and a graph of continuously changing slope is a curve.

Reason (R) is also correct: Non-uniform velocity is defined by a speed that changes over time, which is what bends the graph. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When two cars head toward each other, each sees the other approach at the combined rate, so their relative velocity is the sum of their speeds.

Reason (R) is also correct: Relative velocity is the vector difference of the two velocities; for oppositely directed motions that difference works out to the sum of the magnitudes. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a ball is dropped, it moves under the acceleration due to gravity, so its speed increases continuously until it hits the ground.

Reason (R) is also correct: The acceleration due to gravity acts vertically downward and is constant, so the velocity of the ball increases uniformly as it falls. Hence the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A ball thrown up loses speed, halts for an instant at the top, and then falls back down.

Reason (R) is also correct: The steady downward pull of gravity is what slows the rising ball, stops it, and drives it back. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A cyclist moving in a straight line at constant speed has no change in the magnitude or direction of the velocity, so there is no acceleration.

Reason (R) is also correct: Acceleration occurs only when there is a change in the speed or the direction of motion, and here neither changes. Hence the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A speedometer displays the car's speed at the present moment, which is its instantaneous speed.

Reason (R) is also correct: Instantaneous speed is by definition the speed at one specific moment, matching what the speedometer reads. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Falling freely from the same height, the pin and the ball descend alike and touch down together.

Reason (R) is also correct: The time of fall, $\sqrt{2\text h / \text g}$, has no mass term, so it is the same for both bodies. This is why they land together, and the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a person returns to the starting point, the initial and final positions are the same, so the displacement is zero.

Reason (R) is also correct: Displacement depends only on the initial and final positions and not on the path taken, which is why the displacement is zero for the return journey. Hence the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Constant speed along a straight road means the velocity does not change, so the car has no acceleration.

Reason (R) is also correct: Acceleration is the rate at which velocity changes, and an unchanging velocity gives zero acceleration. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: To a passenger in a moving bus, a fixed tree appears to slide backward.

Reason (R) is also correct: This apparent backward drift arises from the relative motion between the passenger and the tree. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A car braking to a halt is losing speed, and slowing down corresponds to a negative acceleration.

Reason (R) is also correct: A negative acceleration is one that reduces the velocity with time, which is just what happens as the car decelerates. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Moving in opposite directions, each car sees the other approach at the combined rate, so the relative velocity is twice one car's speed.

Reason (R) is also correct: Relative velocity is the vector difference of the two velocities, and for opposite directions that difference adds the magnitudes together. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A body's velocity can be zero at an instant while it is still accelerating — a ball at the top of its flight has zero velocity yet keeps the acceleration due to gravity.

Reason (R) is also correct: When a body reverses its direction it does come momentarily to rest, and at that instant gravity (or whatever force is present) still acts, so it can hold a non-zero acceleration with zero velocity. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Whether a body is dropped or thrown horizontally from the same height, its initial vertical velocity is zero in both cases, so it takes the same time to reach the ground.

Reason (R) is also correct: The horizontal velocity has no effect on the vertical motion, which is governed only by the acceleration due to gravity. Hence the reason correctly explains the assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In the presence of air resistance, the feather is retarded much more than the stone, so the two take different amounts of time to strike the ground.

Reason (R) is false: The force of friction due to air does not depend on the weight of the body. It depends on the shape, size and surface area of the body and on its speed through the air.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Falling freely through the same height, the two masses descend identically and hit the ground at the same moment.

Reason (R) is also correct: The fall time $\sqrt{2\text h / \text g}$ contains no mass, so it is the same regardless of how heavy the body is. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A stationary object keeps the same position at all times, so its position-time graph runs flat, parallel to the time axis.

Reason (R) is also correct: For something that never moves, position does not depend on time, that is, s ≠ f(t), which is exactly what makes the graph horizontal. So the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are false.

Explanation

Assertion (A) is false: A constant speed allows any curved path whatsoever, not merely a circle or a straight line.

Reason (R) is also false: A body can indeed keep a constant speed while accelerating — uniform circular motion is the standard example, where the speed is fixed but the constantly changing direction produces an acceleration.

Therefore, both assertion and reason are false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: When a body returns to its starting point its displacement is zero even though the distance it covered is not zero.

Reason (R) is false: Displacement is the shortest straight-line gap between the initial and final points, not the longer distance.

Therefore, assertion is true but reason is false.

When the distance travelled is directly proportional to the square of time, s ∝ t², the motion follows s = $\frac{1}{2}$at² (starting from rest), which holds only when the acceleration is constant. So the body is moving with uniform acceleration.

When the distance travelled is directly proportional to time, s ∝ t, the body covers equal distances in equal intervals of time, which means the body is moving with uniform velocity.

No. Even when the acceleration is constant in both magnitude and direction, the path of the body is not necessarily a straight line, because the velocity can still change in both magnitude and direction.

For example, a projectile moves under the constant acceleration due to gravity g, yet its path is a parabola, not a straight line.

Both stones strike the ground with the same speed.

The stone thrown up first rises, then falls back through the point of projection, and by the time it returns to that point it is again moving with speed u — now directed downward (air resistance ignored). From that level onward its situation is identical to the stone that was thrown straight down: same downward speed u, same acceleration g, and the same remaining height to fall. With everything matched, they arrive at the earth with equal speed.

The acceleration is zero.

A line running parallel to the time-axis shows a velocity that never changes. Since the acceleration is the slope of that line, and the slope here is zero, the body has no acceleration.

The time of fall depends only on the vertical motion of the body. When the body is dropped, its initial vertical velocity is zero, regardless of whether the train is:

• Stationary
• Moving with a uniform speed v
• Moving with a horizontal acceleration a

The horizontal motion (or horizontal acceleration) of the train does not affect the body's vertical motion. The time of fall is given by

$$\begin{aligned}h = \text{ut} +\frac{1}{2}\text{gt}^2 \\[1em] h = 0 + \frac{1}{2}\text{gt}^2\\[1em] \text{t} = \sqrt{\frac{2\text{h}}{\text{g}}}\end{aligned}$$

which depends only on the height h and the acceleration due to gravity g.

The time of fall is the same whether the train is stationary, moving with uniform speed v, or moving with horizontal acceleration a.

Let the body be projected upward with speed u at a place where gravity is g.

The rise ends where the velocity falls to zero. From v = u - gt with v = 0,

$$\begin{aligned}0 = \text u - \text {gt} \\[1em] \Rightarrow \text t = \frac{\text u}{\text g}\end{aligned}$$

The descent takes an equal time, so the total time for the body to come back is

$\text T = 2\text t = \frac{2\text u}{\text g}$

Because the launch speed u is the same everywhere, the numerator 2u is a fixed number, leaving

$\text T \propto \frac{1}{\text g}$

Hence the time of return varies inversely with the acceleration due to gravity at that place.

B has the greater velocity.

On a time-displacement graph the velocity equals the slope, that is, the tangent of the angle with the time-axis.

$\text v_\text A = \tan 30° = \frac{1}{\sqrt{3}}, \qquad \text v_\text B = \tan 60° = \sqrt{3}$

Taking the ratio,

$\frac{\text v_\text A}{\text v_\text B} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$

So v_A : v_B = 1 : 3, and B is the faster of the two.

In the vertical direction the body simply drops from rest through the height H, so from H = $\frac{1}{2}$gt²,

$\text t = \sqrt{\frac{2\text H}{\text g}}$

Meanwhile it keeps its horizontal velocity v unchanged, so the horizontal distance from the foot of the tower is this velocity times the fall time,

$\text x = \text {vt} = \text v\sqrt{\frac{2\text H}{\text g}}$

Hence the body lands after a time $\sqrt{\frac{2\text H}{\text g}}$, at a distance $\text v\sqrt{\frac{2\text H}{\text g}}$ from the base of the tower.

(i) Distance and displacement

Distance	Displacement
It is the total length of the path covered by a body.	It is the shortest distance between the initial and final positions of the body.
It is a scalar quantity.	It is a vector quantity.
It is always positive for a moving body.	It may be positive, negative or zero.

Example : If a body moves from A (0 m) to B (4 m) and then returns to C (2 m), the distance covered is 4 + 2 = 6 m, whereas the displacement is 2 m.

(ii) Velocity and speed

Speed	Velocity
It is the rate of change of distance with time.	It is the rate of change of displacement with time.
It is a scalar quantity.	It is a vector quantity.
It is always positive.	It may be positive or negative.

Example : A car covering 6 m of road in 3 s has a speed of 2 m s^-1, but if its displacement in that time is 2 m, its velocity is 0.67 m s^-1.

(iii) Velocity-change and acceleration

Velocity-change	Acceleration
It is the difference between the final and the initial velocity.	It is the rate of change of velocity with time.
Its S.I. unit is m s-1.	Its S.I. unit is m s-2.

Example : If the velocity of a body changes from 10 m s^-1 to 20 m s^-1 in 5 s, the velocity-change is 10 m s^-1, whereas the acceleration is 2 m s^-2.

Yes, the displacement can be zero while the distance is not zero. This happens when a body returns to its initial position, for example a body completing one round of a circular track.

No, the reverse is not possible. If the distance covered is zero, the body has not moved at all, so its displacement must also be zero.

Yes, this is possible.

Average velocity divides net displacement by time, whereas average speed divides total path length by time. Whenever a body ends its journey back at the starting point, the displacement — and hence the average velocity — is zero, even though the path length is not.

Example : A man walks 2.5 km from home to the market and comes straight back in 50 min. His displacement is zero, so his average velocity is zero, but he has actually covered 5 km, giving an average speed of 6 km h^-1.

No, the reverse cannot happen. A zero average speed means zero path length, so the body has not moved at all, and then its displacement — and its average velocity — must also be zero.

(i) Yes. A body can have acceleration when its velocity is zero, if it is reversing its direction of motion. For example, in simple harmonic motion the velocity is zero at the extreme position but the acceleration is maximum there; similarly, at the highest point of a body thrown vertically upward the velocity is zero but the body has a downward acceleration.

(ii) Yes. If only the direction of the velocity changes while its magnitude remains constant, the body is accelerated. Uniform circular motion is an example.

(iii) No. If the velocity is uniform, its magnitude and direction both remain unchanged, so the acceleration is zero.

(iv) No. If the displacement is the same in equal intervals of time, the velocity is uniform, and a uniform velocity gives zero acceleration.

(v) Yes. Equal distances in equal intervals of time mean only that the speed is uniform; the direction may still change, as in uniform circular motion, so the body can be accelerated.

(a) Three-dimensional — a kite on a windy day drifts sideways, up and down all at once.

(b) Three-dimensional — the aeroplane can turn and can also climb or descend during its flight.

(c) One-dimensional — the train runs along a single straight track.

(d) Two-dimensional — a fired bullet follows a projectile path in a vertical plane.

(e) Two-dimensional — the coin slides about on the flat board.

(f) Two-dimensional — the earth's orbit lies in a single fixed plane.

(g) Three-dimensional — a boat on a rough sea pitches and rolls in every direction.

(h) One-dimensional — the ball moves straight up and down along one vertical line.

No.

Reading the graph, at the stretch between 2 and 3 hours it would place the car at two different positions for the same instant of time. A single body cannot be in two places at once, and a genuine time-displacement graph can never show two displacement values for one value of time. So this graph cannot represent such a journey.

The velocity of uniform motion equals the slope of the straight x-t line, and the point where the line meets the position axis gives x₀. A positive velocity tilts the line so it makes an acute angle with the time-axis; a negative velocity tilts it the other way, giving an obtuse angle.

Using these rules for the four cases (a), (b), (c) and (d) — with the sign of x₀ fixing the starting intercept and the sign of the velocity fixing the tilt — the graphs come out as shown below.

Because the object keeps a definite, unchanging velocity in one direction, nothing about its velocity varies, so its acceleration is zero.

Accordingly, the time-velocity graph is a horizontal line parallel to the time-axis, while the time-displacement graph is a straight line slanting up from the origin, as shown.

With the acceleration held constant :

• the time-acceleration graph is a horizontal line parallel to the time-axis, since a does not change;
• the time-velocity graph is a slanting straight line, since v = u + at grows steadily;
• the time-displacement graph is a parabola, since s = ut + $\frac{1}{2}$at² is quadratic in time.

These graphs are drawn as shown.

While going to the post-office the person moves slowly, so his velocity is small and positive; while purchasing the postcards he is at rest, so his velocity is zero; while coming back speedily in the opposite direction his velocity is large and negative.

The two graphs are drawn in figures (a) and (b). In figure (a) the areas A and B are equal, because the net displacement of the person is zero.

(i) A frame of reference is a chosen set of coordinate axes together with a clock, used as the standpoint from which motion is described.

(ii) A frame in which Newton's first law correctly describes how a body moves is called an inertial frame of reference.

(iii) A rotating body is always accelerating (its direction of motion keeps turning), so a frame fixed to it does not obey Newton's first law. It is therefore a non-inertial frame of reference.

(i) Between 0 and 15 s the velocity climbs steadily from 0 to 45 m s^-1. The acceleration is the slope of this rising line :

$\text {acceleration (a)} = \frac{\text {AD}}{\text {OD}} = \frac{45\ \text {m s}^{-1}}{15\ \text s} = 3 \text { m s}^{-2}$

(ii) Between 15 and 25 s the velocity holds steady at 45 m s^-1. With no change in velocity,

acceleration (a) = 0

(iii) The displacement over 0 to 25 s is the whole area under the graph, made of the triangle ODA (the rising part) plus the rectangle ABCD (the flat part) :

$$\begin{aligned}\text {Displacement} = \frac{1}{2} \times 15 \times 45 + (45 \times 10) \\[1em] = 337.5 + 450 \\[1em] = 787.5 \text { m}\end{aligned}$$

(i) Graph (a)

Reason — For a body rising under gravity the velocity obeys

v = u - gt ...(i)

which has the straight-line form. So the curved options (b) and (d) are ruled out. Matching (i) with the line y = mx + c, the intercept is u and the slope is - g; that is, the graph must start at a positive value u at t = 0 and then slope downward. Graph (a) is the only one meeting both conditions.

(ii) Graph (c)

Reason — Figure 2 shows the acceleration starting constant and positive, then dropping to zero, and finally turning constant and positive again. The matching velocity graph must therefore rise uniformly (constant acceleration), then stay flat (zero acceleration), then rise uniformly once more. Graph (c) behaves in exactly this way.

(iii) 2 : 1

Reason — On an s-t graph the speed is the slope.

In the first 2 s the slope is

$\text v_{12} = \frac{\text s_0}{2} \text { m s}^{-1}$

Over the next 4 s (from 2 s to 6 s) the slope is

$\text v_{26} = \frac{\text s_0}{4} \text { m s}^{-1}$

Their ratio is

$\frac{\text v_{12}}{\text v_{26}} = \frac{\text s_0 / 2}{\text s_0 / 4} = \frac{2}{1} = 2 : 1$

(iv) Graph (c)

Reason — For uniformly accelerated motion,

v = u + at

Starting from rest makes u = 0, so v = at. This is of the form y = mx with zero intercept and positive slope $a$. A straight line rising from the origin. Graph (c) shows exactly this.

(v) Graph (b)

Reason — Linking speed to distance for a body starting from rest,

v² = u² + 2as

with u = 0 gives v² = 2as. This matches the parabola y² = 4ax opening along the X-axis, so the speed-distance curve is a parabola. Option (b) is correct.

Consider an object travelling in a straight line at a uniform velocity. Let its position be s₁ at time t₁ and s₂ at time t₂. Because the velocity does not change, plotting position against time gives a straight line, as shown.

By its very definition,

$$\begin{aligned}\text {velocity} = \frac{\text {change in position}}{\text {time taken}} \\[1em] = \frac{\text s_2 - \text s_1}{\text t_2 - \text t_1}\end{aligned}$$

But the ratio $\frac{\text s_2 - \text s_1}{\text t_2 - \text t_1}$ is nothing other than the slope of the straight line — the tangent of the angle θ it makes with the time-axis.

So the slope of the time-displacement graph equals the velocity of the object.

Negative velocity : A velocity that comes out negative simply means the object is heading opposite to the direction we have called positive. On the graph the line then leans the other way, making an obtuse angle with the time-axis, so its slope is negative.

Example : If a car going east is taken to have positive velocity, then an identical car going west along the same road has a negative velocity.

Consider an object moving straight with a constant acceleration a. Let its velocity be u at t = 0 and v at time t. A constant acceleration makes the velocity rise linearly, so the time-velocity graph is the straight line AB shown below.

The distance covered is the area between this line and the time-axis, that is, the area of the trapezium OABD, which we split into a rectangle and a triangle :

$$\begin{aligned}\text {Area of trapezium OABD} = \text {Area of rectangle OACD} + \text {Area of }\Delta \text {ABC} \\[1em] = (\text {OA} \times \text {OD}) + \frac{1}{2}(\text {AC} \times \text {BC}) \\[1em] = \text {ut} + \frac{1}{2}\text t(\text v - \text u)\end{aligned}$$

For constant acceleration, v - u = at, so

$$\begin{aligned}\text {Area} = \text {ut} + \frac{1}{2}\text t(\text {at}) \\[1em] = \text {ut} + \frac{1}{2}\text {at}^2\end{aligned}$$

This is exactly the formula for the distance s covered in time t.

Hence the area enclosed by the time-velocity graph and the time-axis equals the distance travelled in that interval.

Consider an object with a constant acceleration $a$ acting over a time-interval Δt. Since $a$ does not change, the time-acceleration graph is a horizontal line parallel to the time-axis, as shown.

The area shut in between this line and the time-axis is simply the area of a rectangle :

Area = a x (t₂ - t₁) = aΔt

Now, from the definition of acceleration,

$$\begin{aligned}\text a = \frac{\text v - \text u}{\text t_2 - \text t_1} = \frac{\text v - \text u}{Δ\text t} \\[1em] \Rightarrow \text v - \text u = \text {aΔt}\end{aligned}$$

where u and v are the velocities at the start and end of the interval.

So the rectangular area aΔt is precisely (v - u), the change in velocity over that interval.

Hence the area enclosed by the time-acceleration graph and the time-axis equals the velocity-change for that interval.

Consider an object moving straight with a constant acceleration $a$. Let $u$ be its velocity at $t$ = 0 and $v$ its velocity at time $t$, and let $s$ be the distance it covers in that time. Since $a$ is constant, the time-velocity graph is the straight line AB shown below.

(a) Deriving v = u + at

Acceleration is the rate of change of velocity, which is the slope of the line AB :

$$\begin{aligned}\text a = \frac{\text{CB}}{\text{AC}} = \frac{\text v - \text u}{\text t} \\[1em] \Rightarrow \text {at} = \text v - \text u \\[1em] \Rightarrow \text v = \text u + \text {at}\end{aligned}$$

(b) Deriving s = ut + $\frac{1}{2}$at²

The distance is the area under the graph, that is, the trapezium OABD split as a rectangle plus a triangle :

$$\begin{aligned}\text s = \text {Area of rectangle OACD} + \text {Area of }\Delta \text {ABC} \\[1em] = (\text {DC} \times \text {AC}) + \frac{1}{2}(\text {AC} \times {CB})\\[1em] = (\text u \times \text t) + \frac{1}{2}(\text t \times (\text v - \text u))\end{aligned}$$

Substituting v - u = at from part (a),

$$\begin{aligned}\text s = \text {ut} + \frac{1}{2}\text t(\text {at}) \\[1em] = \text {ut} + \frac{1}{2}\text {at}^2\end{aligned}$$

(c) Deriving v² = u² + 2as

The same area of the trapezium OABD can be written using the two parallel sides :

$$\begin{aligned}\text s = \frac{1}{2}(\text {OA} + \text {DB}) \times \text {OD} \\[1em] = \frac{1}{2}(\text u + \text v) \times \text t\end{aligned}$$

From part (a), $\text t = \frac{\text v - \text u}{\text a}$. Substituting,

$$\begin{aligned}\text s = \frac{1}{2}(\text u + \text v) \times \frac{\text v - \text u}{\text a} \\[1em] = \frac{\text v^2 - \text u^2}{2\text a} \\[1em] \Rightarrow 2\text {as} = \text v^2 - \text u^2 \\[1em] \Rightarrow \text v^2 = \text u^2 + 2\text {as}\end{aligned}$$

Relative velocity : The relative velocity of one body with respect to another is the rate at which the first body's position changes as judged from the second body. In other words, it is how the first body's motion looks to an observer riding on the second body.

Suppose two particles, 1 and 2, travel along a straight line (the X-axis) with uniform velocities v₁ and v₂ measured from the ground. Then the velocity of particle 2 as seen from particle 1 is

$\vec{\text v}_{21} = \vec{\text v}_2 - \vec{\text v}_1$

and the velocity of particle 1 as seen from particle 2 is

$\vec{\text v}_{12} = \vec{\text v}_1 - \vec{\text v}_2$

Special cases :

• If both the particles move in the same direction, the magnitude of the relative velocity is the difference of their speeds, that is, v₁ - v₂. If v₁ = v₂, the relative velocity is zero and the two particles stay at a constant distance apart.
• If the two particles move in opposite directions, the magnitude of the relative velocity is the sum of their speeds, that is, v₁ + v₂.

(i) The speed is given by the slope of the t-s graph. The slope is maximum for the steepest portion of the graph, in which the cyclist covers 20 km in 2 h.

$\text {Maximum speed} = \frac{20\ \text {km}}{2\ \text h} = 10 \text { km h}^{-1}$

(ii) Over the entire ride the cyclist travels 35 km in a total of 5 h, so

$\text {Average speed} = \frac{\text {total distance}}{\text {total time}} = \frac{35\ \text {km}}{5\ \text h} = 7 \text { km h}^{-1}$

The time-velocity graph is drawn as shown.

(i) The body moves fastest between 0 and 1 second, since the time-displacement graph is steepest there and a steeper slope means a larger velocity.

(ii) Adding up the motion, the body goes 20 m forward and 20 m backward, so the total distance travelled is 40 metre.

(iii) As the body finishes exactly where it began, its net displacement is zero.

(iv) Up to A the graph is a straight line, so the velocity is unchanging over that stretch. An unchanging velocity means the acceleration from the beginning up to A is zero.

Given,

• First leg : velocity v₁ = 20 km h^-1 for time t₁ = 0.1 h
• Second leg : velocity v₂ = 60 km h^-1 for time t₂ = 0.4 h
• Third leg : velocity v₃ = 20 km h^-1 for time t₃ = 0.2 h

The time-velocity graph for the car is drawn as shown.

(i) On a velocity-time graph the distance is the area beneath it. For the first leg this is the rectangle of height v₁ and width t₁ :

$\text s_1 = \text v_1 \times \text t_1 = 20 \times 0.1 = 2 \text { km}$

(ii) The whole journey is the sum of the three rectangular areas :

$$\begin{aligned}\text s = (\text v_1\text t_1) + (\text v_2\text t_2) + (\text v_3\text t_3) \\[1em] = (20 \times 0.1) + (60 \times 0.4) + (20 \times 0.2) \\[1em] = 2 + 24 + 4 = 30 \text { km}\end{aligned}$$

(iii) In the first leg the car covers 2 km, so the car still needs 14 - 2 = 12 km, which it covers in the second leg at 60 km h^-1 :

$\text {Time for 12 km} = \frac{12\ \text {km}}{60\ \text {km h}^{-1}} = 0.2 \text { h}$

Adding the first leg's 0.1 h,

$\text {Total time for 14 km} = 0.1 + 0.2 = 0.3 \text { h}$

On the graph, the area up to t = 0.3 h (shaded) represents this first 14 km.

Given,

• Initial velocity, u = 0 (starts from rest)
• Acceleration, a = 8.0 m s^-2
• Time, t = 5 s

Since the velocity grows steadily from zero, the time-velocity graph is a straight line through the origin, as shown.

(i) Using v = u + at,

$\text v = 0 + (8.0 \times 5) = 40 \text { m s}^{-1}$

(ii) The distance is the area under the line — a triangle of base OB and height BA :

$$\begin{aligned}\text s = \frac{1}{2} \times \text {OB} \times \text {BA} \\[1em] = \frac{1}{2} \times 5 \times 40 \\[1em] = 100 \text { m}\end{aligned}$$

(iii) The average velocity is the displacement divided by the time :

$\text v_\text{avg} = \frac{\text {Total displacement}}{\text {Total time}} = \frac{100\ \text m}{5\ \text s} = 20 \text { m s}^{-1}$

Given (read from the time-velocity graph),

• The speed climbs from 0 to 8 m s^-1 between t = 0 and t = 2 s, stays at 8 m s^-1 up to t = 10 s, drops to 4 m s^-1 between t = 10 s and t = 12 s, and holds at 4 m s^-1 up to t = 16 s.

Distance in 16 s. The distance is the whole area beneath the graph :

$$\begin{aligned}\text s = \frac{1}{2}(2 \times 8) + (10 - 2) \times 8 + \frac{1}{2}(8 + 4)(12 - 10) + (16 - 12) \times 4 \\[1em] = 8 + 64 + 12 + 16 \\[1em] = 100 \text { m}\end{aligned}$$

Acceleration at t = 11 s. At this instant the graph is on its downward slope from 8 m s^-1 (at 10 s) to 4 m s^-1 (at 12 s). The acceleration is the slope at 11 sec :

$\text a = \frac{4 - 8}{12 - 10} = \frac{-4}{2} = -2 \text { m s}^{-2}$

The minus sign shows the runner is slowing down at that moment.

Given,

• Acceleration during powered flight, a = 20 m s^-2
• Duration of powered flight, t₁ = 1.0 min = 60 s
• Initial velocity, u = 0
• g = 10 m s^-2

Powered stage. The velocity gained by the time the fuel runs out, from v = u + at₁ :

$\text v = \text u + \text {at}_1 = 0 + (20 \times 60) = 1200 \text { m s}^{-1}$

The height climbed in this stage, from h₁ = ut₁ + $\frac{1}{2}$at₁² :

$$\begin{aligned}\text h_1 = \text {ut}_1 + \frac{1}{2}\text a\text t_1^2 \\[1em] = (0 \times 60) + \frac{1}{2} \times 20 \times (60)^2 \\[1em] = 10 \times 3600 \\[1em] = 36000 \text { m}\end{aligned}$$

Free rise after the fuel ends. Now the rocket coasts upward with initial velocity 1200 m s^-1 against gravity (retardation g = 10 m s^-2), stopping at the top. From v₂² = u₂² - 2gh₂ with v₂ = 0 :

$$\begin{aligned}0 = (1200)^2 - 2\text {gh}_2 \\[1em] \Rightarrow \text h_2 = \frac{(1200)^2}{2 \times 10} = 72000 \text { m}\end{aligned}$$

So the greatest height is

$\text H = \text h_1 + \text h_2 = 36000 + 72000 = 108000 \text { m} = 1.08 \times 10^5 \text { m}$

Time for this coasting rise, from v₂ = u₂ - gt₂ with v₂ = 0 :

$$\begin{aligned}0 = 1200 - 10\ \text t_2 \\[1em] \Rightarrow \text t_2 = \frac{1200}{10} = 120 \text { s}\end{aligned}$$

Falling stage. From the peak the rocket drops freely through 108000 m, starting from rest. From h = $\frac{1}{2}$gt₃² :

$$\begin{aligned}\text t_3 = \sqrt{\frac{2\text H}{\text g}} \\[1em] = \sqrt{\frac{2 \times 108000}{10}} \\[1em] = \sqrt{21600} \\[1em] = 147 \text { s}\end{aligned}$$

Total time before landing :

$\text T = \text t_1 + \text t_2 + \text t_3 = 60 + 120 + 147 = 327 \text { s}$

(i) Reading the graph, the speed falls steadily to zero in the first 4 s and then rises steadily again — so the motion is uniformly retarded while going up (first 4 s) and uniformly accelerated while coming down.

(ii) The graph meets the velocity-axis at 40, so the initial velocity is

$\text u = 40 \text { m s}^{-1}$

The acceleration is the slope of the line :

$\text a = \frac{0 - 40}{4 - 0} = -10 \text { m s}^{-2}$

(iii) The greatest height is the area under the graph during the upward phase (the triangle over the first 4 s) :

$\text h = \frac{1}{2} \times 4 \times 40 = 80 \text { m}$

(iv) The graph shows the body back at the launch level at t = 8 s, so it returns after 8 s.

Given,

• Distance between the towns, D = 480 km
• Time for the first car (A to B), t_A = 8 h
• Time for the second car (B to A), t_B = 12 h

Speed of the first car :

$\text v_1 = \frac{\text D}{\text t_\text A} = \frac{480}{8} = 60 \text { km h}^{-1}$

Speed of the second car :

$\text v_2 = \frac{\text D}{\text t_\text B} = \frac{480}{12} = 40 \text { km h}^{-1}$

The cars head toward each other, so they close the gap at their combined speed :

$\text v_\text{rel} = \text v_1 + \text v_2 = 60 + 40 = 100 \text { km h}^{-1}$

∴ time taken to meet:

$\text t = \frac{\text D}{\text v_\text{rel}} = \frac{480}{100} = 4.8 \text { h}$

Distance of the meeting point from town A (using the first car's speed) :

$\text s = \text v_1 \times \text t = 60 \times 4.8 = 288 \text { km}$

Hence the two cars meet after 4.8 h, 288 km from town A.

(i) The acceleration between A and B is the slope of that segment. Reading the co-ordinates of A and B from the graph,

$\text a = \frac{2 - 5}{3 - 2} = \frac{-3}{1} = -3 \text { m s}^{-2}$

The minus sign tells us the car is slowing down over this portion.

(ii) The distance in the last 4 s is the area under the graph there. That region is a trapezium with parallel sides (7 - 3) and (5 - 3) and width 2 :

$$\begin{aligned}\text s = \frac{1}{2} \times [(7 - 3) + (5 - 3)] \times 2 \\[1em] = \frac{1}{2} \times (4 + 2) \times 2 \\[1em] = 6 \text { m}\end{aligned}$$

Hence the acceleration from A to B is - 3 m s^-2 and the distance in the last 4 s is 6 m.

Given,

• Initial velocity, u = 0 (starts from rest)
• Final velocity, v = 180 km h^-1 = $\frac{180 \times 1000}{3600}$ = 50 m s^-1
• Time, t = 25 s

The acceleration is uniform, so the average velocity is the mean of the start and end speeds, and distance = average velocity × time :

$$\begin{aligned}\text s = \left(\frac{\text u + \text v}{2}\right)\text t \\[1em] = \left(\frac{0 + 50}{2}\right) \times 25 \\[1em] = 25 \times 25 \\[1em] = 625 \text { m}\end{aligned}$$

Hence the car covers 625 m in this interval.

Given,

• Initial velocity, u = 20 m s^-1 (upward)
• Total time of flight, t = 5 s
• g = 9.8 m s^-2

Take upward as positive. The ball's displacement when it lands (measured from the top of the tower) follows from s = ut - $\frac{1}{2}$gt² :

$$\begin{aligned}\text s = (20 \times 5) - \frac{1}{2} \times 9.8 \times (5)^2 \\[1em] = 100 - 122.5 \\[1em] = -22.5 \text { m}\end{aligned}$$

The negative result means the landing point is 22.5 m below the launch point, so the tower is 22.5 m high.

Velocity on striking the ground, from v = u - gt :

$$\begin{aligned}\text v = 20 - (9.8 \times 5) \\[1em] = 20 - 49 \\[1em] = -29 \text { m s}^{-1}\end{aligned}$$

The minus sign shows it is moving downward.

Hence the tower is 22.5 m high and the ball hits the ground at 29 m s^-1 (downward).

Given,

• Initial velocity, u = 500 cm s^-1 = 5 m s^-1 (downward)
• Height of the building, h = 30 m
• g = 9.8 m s^-2

Take downward as positive.

(a) Velocity after 0.5 s, from v = u + gt :

$$\begin{aligned}\text v = 5 + (9.8 \times 0.5) \\[1em] = 5 + 4.9 \\[1em] = 9.9 \text { m s}^{-1}\end{aligned}$$

(b) Position after 1.5 s, from s = ut + $\frac{1}{2}$gt² :

$$\begin{aligned}\text s = (5 \times 1.5) + \frac{1}{2} \times 9.8 \times (1.5)^2 \\[1em] = 7.5 + 11.025 \\[1em] = 18.525 \text { m}\end{aligned}$$

So the ball is 18.525 m below the roof at this instant.

(c) Speed on striking the ground, from v² = u² + 2gh :

$$\begin{aligned}\text v^2 = (5)^2 + 2 \times 9.8 \times 30 \\[1em] = 25 + 588 \\[1em] = 613 \\[1em] \Rightarrow \text v = 24.76 \text { m s}^{-1}\end{aligned}$$

Hence the ball hits the earth at 24.76 m s^-1.

Given,

• Speed downstream (with the current), swimming with the stream = 25 km h^-1
• Speed upstream (against the current), swimming against the stream = 15 km h^-1

Let the swimmer's speed in still water be v_p and the river's speed be v_r.

Going with the stream, the current helps, so the two speeds add :

$\text v_\text p + \text v_\text r = 25 \quad \text{...(i)}$

Going against the stream, the current hinders, so they subtract :

$\text v_\text p - \text v_\text r = 15 \quad \text{...(ii)}$

Adding (i) and (ii) removes v_r :

$$\begin{aligned}2\text v_\text p = 40 \\[1em] \Rightarrow \text v_\text p = 20 \text { km h}^{-1}\end{aligned}$$

Subtracting (ii) from (i) removes v_p :

$$\begin{aligned}2\text v_\text r = 10 \\[1em] \Rightarrow \text v_\text r = 5 \text { km h}^{-1}\end{aligned}$$

Hence the swimmer's still-water speed is 20 km h^-1 and the river flows at 5 km h^-1.

Given,

• Speed of train A, v_A = 60 km h^-1
• Speed of train B, v_B = 45 km h^-1

(i) Both moving the same way — take that direction as positive :

$$\begin{aligned}\vec{\text v}_\text{AB} = \vec{\text v}_\text A - \vec{\text v}_\text B \\[1em] = 60 - 45 \\[1em] = 15 \text { km h}^{-1}\end{aligned}$$

(ii) Moving in opposite directions — now v_B is negative, v_B = - 45 km h^-1 :

$$\begin{aligned}\vec{\text v}_\text{AB} = \vec{\text v}_\text A - \vec{\text v}_\text B \\[1em] = 60 - (-45) \\[1em] = 105 \text { km h}^{-1}\end{aligned}$$

Hence A moves at 15 km h^-1 relative to B when they travel the same way, and at 105 km h^-1 when they travel opposite ways.

(i) At t = 0, car A's graph starts at 5 km while car B's starts from the origin, so A begins 5 km ahead of B.

(ii) On a distance-time graph the speed is the slope. Reading the graphs,

$\text {Speed of B} = \frac{20 - 0}{3 - 0} = \frac{20}{3} \text { km h}^{-1}$

$\text {Speed of A} = \frac{10 - 0}{3 - 0} = \frac{10}{3} \text { km h}^{-1}$

Because B's line is steeper, car B is the faster one.

(iii) The two lines cross where B overtakes A. From the graph this intersection is at t = 1.5 h and s = 10 km.

Hence car B catches car A after 1.5 h, at a point 10 km from the start.

On a time-acceleration graph, the change in velocity across any interval equals the area under the graph for that interval. Building up the velocity from v = 0 at t = 0 :

• Where the acceleration is constant and positive, the velocity climbs steadily, so the v-t graph is a straight line sloping upward.
• Where the acceleration is zero, the velocity holds steady, so the v-t graph runs flat (parallel to the time-axis).
• Where the acceleration is constant and negative, the velocity falls steadily, so the v-t graph is a straight line sloping downward.

Joining these points gives the time-velocity graph shown in Fig. (b).

Given,

• Height of the house, h = 100 m
• g = 9.8 m s^-2

The upward-thrown ball only just reaches the roof, so its velocity there is zero. From v² = u² - 2gh with v = 0, its launch speed is

$$\begin{aligned}\text u^2 = 2\text {gh} \\[1em] \Rightarrow \text u = \sqrt{2 \times 9.8 \times 100} \\[1em] = \sqrt{1960} \\[1em] = 44.27 \text { m s}^{-1}\end{aligned}$$

Let the balls meet after time t, with heights measured from the ground.

Height of the ball thrown up : $\text x_1 = \text {ut} - \frac{1}{2}\text {gt}^2$

Height of the ball dropped from the roof : $\text x_2 = 100 - (0\times \text{t} +\frac{1}{2}\text {gt}^2)$

At the meeting point x₁ = x₂. The $\frac{1}{2}$gt² terms cancel :

$$\begin{aligned}\text {ut} - \frac{1}{2}\text {gt}^2 = 100 - \frac{1}{2}\text {gt}^2 \\[1em] \Rightarrow \text {ut} = 100 \\[1em] \Rightarrow \text t = \frac{100}{44.27} = 2.26 \text { s}\end{aligned}$$

Substituting this time into the height of the dropped ball,

$$\begin{aligned}\text x = 100 - \frac{1}{2} \times 9.8 \times (2.26)^2 \\[1em] = 100 - 25 \\[1em] = 75 \text { m}\end{aligned}$$

Hence the two balls cross each other 75 m above the ground.

Given,

• Launch speed of ball A, u_A = 20 m s^-1 (upward, from the ground)
• Launch speed of ball B, u_B = 20 m s^-1 (downward, from a height of 40 m)
• g = 9.8 m s^-2
• Initial separation = 40 m

Both balls share the same downward acceleration g, so relative to each other there is no acceleration. Their relative velocity therefore stays fixed at the sum of the launch speeds :

$\text v_\text{rel} = 20 + 20 = 40 \text { m s}^{-1}$

Time to cover the 40 m gap :

$\text t = \frac{40\ \text m}{40\ \text {m s}^{-1}} = 1 \text { s}$

So, both balls will collide after 1 second.

Now find where they are at t = 1 s. Height of ball A above the ground, x = u_At - $\frac{1}{2}$gt² :

$$\begin{aligned}\text x = (20 \times 1) - \frac{1}{2} \times 9.8 \times (1)^2 \\[1em] = 20 - 4.9 \\[1em] = 15.1 \text { m}\end{aligned}$$

Hence the balls collide 1 s after launch, 15.1 m above the ground.

Given,

• Initial velocity, u = 10 m s^-1 (upward)
• Height of the building, h = 40 m
• g = 10 m s^-2

Take upward as positive. On reaching the ground the stone's displacement from the top is s = - 40 m.

(a) Time to hit the ground, from s = ut - $\frac{1}{2}$gt² :

$$\begin{aligned}-40 = 10\ \text t - 5\ \text t^2 \\[1em] \Rightarrow 5\ \text t^2 - 10\ \text t - 40 = 0 \\[1em] \Rightarrow \text t^2 - 2\ \text t - 8 = 0 \\[1em] \Rightarrow (\text t - 4)(\text t + 2) = 0 \\[1em] \Rightarrow \text t = 4 \text { s}\end{aligned}$$

(the negative root being rejected).

(b) It returns to the launch point when its displacement is again zero. Setting s = 0 in s = ut₁ - $\frac{1}{2}$gt₁² :

$$\begin{aligned}0 = \text {ut}_1 - \frac{1}{2}\text {gt}_1^2 \\[1em] \Rightarrow \text {ut}_1 = \frac{1}{2}\text {gt}_1^2 \\[1em] \Rightarrow \text t_1 = \frac{2\text u}{\text g} = \frac{2 \times 10}{10} = 2 \text { s}\end{aligned}$$

(c) Velocity just before hitting the ground, from v = u - gt with t = 4 s :

$$\begin{aligned}\text v = 10 - (10 \times 4) \\[1em] = 10 - 40 \\[1em] = -30 \text { m s}^{-1}\end{aligned}$$

The minus sign marks it as downward.

Hence the stone lands after 4 s at 30 m s^-1 (downward), having passed the launch point after 2 s.

Given,

• Free-fall distance before the parachute opens, h = 40 m
• Initial velocity, u = 0
• Retardation after the parachute opens, a = 2.0 m s^-2
• Landing velocity, v₁ = 3.0 m s^-1
• g = 9.8 m s^-2

Stage 1 — free fall through 40 m. Speed at the end, from v² = u² + 2gh :

$$\begin{aligned}\text v^2 = 0 + 2 \times 9.8 \times 40 = 784 \\[1em] \Rightarrow \text v = 28 \text { m s}^{-1}\end{aligned}$$

Time for this fall, from v = u + gt₁ :

$\text t_1 = \frac{\text v}{\text g} = \frac{28}{9.8} = 2.86 \text { s}$

Stage 2 — after the parachute opens. Here the initial speed is u₁ = 28 m s^-1, the final speed is v₁ = 3.0 m s^-1, and the retardation is a = 2.0 m s^-2. Distance covered, from v₁² = u₁² - 2as :

$$\begin{aligned}(3)^2 = (28)^2 - 2 \times 2 \times \text s \\[1em] \Rightarrow 9 = 784 - 4\ \text s \\[1em] \Rightarrow \text s = \frac{775}{4} = 193.75 \text { m}\end{aligned}$$

Time for this stage, from v₁ = u₁ - at₂ :

$\text t_2 = \frac{\text u_1 - \text v_1}{\text a} = \frac{28 - 3}{2} = 12.5 \text { s}$

Combining the two stages.

Height of the aeroplane = 40 + 193.75 = 233.75 ≈ 234 m

Total time in the air = t₁ + t₂ = 2.86 + 12.5 = 15.4 s

Given,

• Initial velocity of the parachutist, u₁ = 10 m s^-1 (upward, sharing the balloon's motion)
• Starting height = 45 m
• Free-fall time before the parachute opens = 3 s
• Retardation after the parachute opens = 5 m s^-2
• g = 10 m s^-2

Take upward as positive.

(a) Displacement during the first 3 s, from s = u₁t₁ - $\frac{1}{2}$gt₁² :

$$\begin{aligned}\text s = (10 \times 3) - \frac{1}{2} \times 10 \times (3)^2 \\[1em] = 30 - 45 = -15 \text { m}\end{aligned}$$

So he drops 15 m below his exit point, and his height above the ground is

$45 - 15 = 30 \text { m}$

(b) In those same 3 s the balloon keeps rising at 10 m s^-1, reaching

$45 + (10 \times 3) = 75 \text { m}$

So the parachutist is now

$75 - 30 = 45 \text { m}$

below the balloon.

(c) His velocity when the parachute opens, from v₁ = u₁ - gt₁ :

$\text v_1 = 10 - (10 \times 3) = -20 \text { m s}^{-1}$

that is, 20 m s^-1 downward. He then covers the remaining 30 m with a 5 m s^-2 retardation. From v₂² = u₂² - 2as (taking downward magnitudes, u₂ = 20 m s^-1) :

$$\begin{aligned}\text v_2^2 = (20)^2 - 2 \times 5 \times 30 \\[1em] = 400 - 300 = 100 \\[1em] \Rightarrow \text v_2 = 10 \text { m s}^{-1}\end{aligned}$$

(d) Time for this final 30 m, from v₂ = u₂ - at₂ :

$\text t_2 = \frac{\text u_2 - \text v_2}{\text a} = \frac{20 - 10}{5} = 2 \text { s}$

So the total time from leaving the balloon to landing is

$\text t_1 + \text t_2 = 3 + 2 = 5 \text { s}$

Hence: (a) 30 m above the ground, (b) 45 m below the balloon, (c) 10 m s^-1, (d) 5 s.

Given,

• Acceleration during powered flight, a = 10 m s^-2
• Initial velocity, u = 0
• Duration of powered flight, t₁ = 1 min = 60 s
• g = 10 m s^-2

Velocity of the rocket when the fuel runs out, from v = u + at₁ :

$\text v = 0 + (10 \times 60) = 600 \text { m s}^{-1}$

After that the rocket coasts upward against gravity, so it is retarded at 10 m s^-2 until its velocity reaches zero at the top. From v' = v - gt₂ with v' = 0 :

$$\begin{aligned}0 = 600 - 10\ \text t_2 \\[1em] \Rightarrow \text t_2 = \frac{600}{10} = 60 \text { s} = 1 \text { min}\end{aligned}$$

Hence the highest point is reached 1 min after the fuel is exhausted.

Given,

• Length of each train = 200 m
• Time to overtake (same direction) = 40 s
• Time to cross (opposite directions) = 20 s

Let the speeds be v_A and v_B, with v_A > v_B. In either case one train must clear a total length of

$200 + 200 = 400 \text { m}$

Overtaking (same direction). The relative velocity is v_A - v_B, and it clears 400 m in 40 s :

$$\begin{aligned}\frac{400}{\text v_\text A - \text v_\text B} = 40 \\[1em] \Rightarrow \text v_\text A - \text v_\text B = 10 \quad \text{...(i)}\end{aligned}$$

Crossing (opposite directions). The relative velocity is v_A + v_B, clearing 400 m in 20 s :

$$\begin{aligned}\frac{400}{\text v_\text A + \text v_\text B} = 20 \\[1em] \Rightarrow \text v_\text A + \text v_\text B = 20 \quad \text{...(ii)}\end{aligned}$$

Adding (i) and (ii),

$2\text v_\text A = 30 \Rightarrow \text v_\text A = 15 \text { m s}^{-1}$

Then from (ii),

$\text v_\text B = 20 - 15 = 5 \text { m s}^{-1}$

Hence the trains travel at 15 m s^-1 and 5 m s^-1.

Given,

• Initial velocity of the packet, u = 12 m s^-1 (upward, sharing the balloon's motion)
• Height above the earth, h = 65 m
• g = 10 m s^-2

Take upward as positive. When the packet reaches the earth its displacement is s = - 65 m. From s = ut - $\frac{1}{2}$gt² :

$$\begin{aligned}-65 = 12\ \text t - \frac{1}{2} \times 10 \times \text t^2 \\[1em] \Rightarrow -65 = 12\ \text t - 5\ \text t^2 \\[1em] \Rightarrow 5\ \text t^2 - 12\ \text t - 65 = 0\end{aligned}$$

Solving this quadratic,

$$\begin{aligned}\text t = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 5 \times (-65)}}{2 \times 5} \\[1em] = \frac{12 \pm \sqrt{144 + 1300}}{10} \\[1em] = \frac{12 \pm 38}{10}\end{aligned}$$

Taking the positive root,

$\text t = \frac{12 + 38}{10} = \frac{50}{10} = 5 \text { s}$

Hence the packet reaches the earth after 5 s.

6.25 ms^-2

Reason — The distance covered in a particular second is given by

$\text S_\text n = \text u + \frac{1}{2}\text a(2\text n - 1)$

Applying it to the n^th second,

$102.5 = \text u + \frac{1}{2}\text a(2\text n - 1) \quad \text{...(i)}$

and to the (n + 2)^th second,

$$\begin{aligned}115.0 = \text u + \frac{1}{2}\text a{2(\text n + 2) - 1} \\[1em] = \text u + \frac{1}{2}\text a(2\text n + 3) \quad \text{...(ii)}\end{aligned}$$

Subtracting (i) from (ii) cancels u and n :

$$\begin{aligned}115.0 - 102.5 = \frac{1}{2}\text a[(2\text n + 3) - (2\text n - 1)] \\[1em] 12.5 = \frac{1}{2}\text a \times 4 \\[1em] 12.5 = 2\text a \\[1em] \Rightarrow \text a = 6.25 \text { m s}^{-2}\end{aligned}$$

Reason —

(a) Position-time graph (correct). The curve is a downward-opening parabola: the position first grows, tops out, and then falls. Since the slope of an x-t graph is the velocity, that slope is positive at first, eases to zero at the peak, and then turns negative — exactly what a body under a steady negative acceleration does as it slows, halts for an instant, and reverses.

(b) Velocity-time graph (correct). This is a straight line of constant negative slope, i.e. uniform negative acceleration. The velocity falls off linearly, hits zero once, and then changes sign — the body moves forward, decelerates, stops briefly, and then speeds up the other way. A constant-acceleration motion is faithfully shown.

(c) Velocity-time graph (incorrect). Here the curve is drawn as a sideways parabola that loops back toward the time axis, so it is not single-valued in time: a vertical line at most instants meets it at two points, assigning two velocities to the same moment. A body can hold only one velocity at any instant, so this graph is invalid — which is why (c) is the odd one out.

(d) Distance-time graph (correct). Distance is a scalar that can only build up, and the graph rises throughout. Its slope shrinks as the body slows, flattens near the turning instant, and then steepens again once the body moves off in the reverse direction — consistent with the other correct graphs.

$\text a + \frac{\text b^2}{3\text c}$

Reason — The position is x(t) = at + bt² - ct³.

Differentiating once gives the velocity :

$\text v = \frac{\text {dx}}{\text {dt}} = \text a + 2\text {bt} - 3\text {ct}^2$

Differentiating again gives the acceleration :

$\text A = \frac{\text {dv}}{\text {dt}} = 2\text b - 6\text {ct}$

Setting the acceleration to zero to find the required instant :

$$\begin{aligned}2\text b - 6\text {ct} = 0 \\[1em] \Rightarrow \text t = \frac{\text b}{3\text c}\end{aligned}$$

Putting this t back into the velocity :

$$\begin{aligned}\text v = \text a + 2\text b\left(\frac{\text b}{3\text c}\right) - 3\text c\left(\frac{\text b}{3\text c}\right)^2 \\[1em] = \text a + \frac{2\text b^2}{3\text c} - \frac{\text b^2}{3\text c} \\[1em] = \text a + \frac{\text b^2}{3\text c}\end{aligned}$$

50

Reason — The position vector is $\vec{\text r}(\text t) = 15\text t^2\ \hat{\text i} + (4 - 20\text t^2)\ \hat{\text j}$.

Differentiating once gives the velocity :

$\vec{\text v} = \frac{\text d\vec{\text r}}{\text {dt}} = 30\text t\ \hat{\text i} - 40\text t\ \hat{\text j}$

Differentiating again gives the acceleration :

$\vec{\text a} = \frac{\text d\vec{\text v}}{\text {dt}} = 30\ \hat{\text i} - 40\ \hat{\text j}$

This acceleration is the same at all times, so its magnitude at t = 1 is

$$\begin{aligned}|\vec{\text a}| = \sqrt{(30)^2 + (-40)^2} \\[1em] = \sqrt{900 + 1600} \\[1em] = \sqrt{2500} = 50\end{aligned}$$

$\frac{\text b^2 \tau}{2}$

Reason — The speed is given as $\text v = \text b\sqrt{\text x}$.

To get the acceleration, write it using the chain rule :

$\text a = \frac{\text {dv}}{\text {dt}} = \frac{\text {dv}}{\text {dx}} \times \frac{\text {dx}}{\text {dt}} = \text v\frac{\text {dv}}{\text {dx}}$

Differentiating v with respect to x :

$\frac{\text {dv}}{\text {dx}} = \frac{\text b}{2\sqrt{\text x}}$

So the acceleration is

$\text a = \text b\sqrt{\text x} \times \frac{\text b}{2\sqrt{\text x}} = \frac{\text b^2}{2}$

which is a constant. The particle starts at x = 0 where v = 0, so applying v = u + at with u = 0 and t = τ :

$\text v = \frac{\text b^2}{2}\tau = \frac{\text b^2 \tau}{2}$

$20\sqrt{2}$ m

Reason —

Given,

• Initial position, $\vec{\text r}_0 = (2\ \hat{\text i} + 4\ \hat{\text j})$ m
• Initial velocity, $\vec{\text u} = (5\ \hat{\text i} + 4\ \hat{\text j})$ m s^-1
• Acceleration, $\vec{\text a} = (4\ \hat{\text i} + 4\ \hat{\text j})$ m s^-2
• Time, t = 2 s

The position at time t follows from $\vec{\text r} = \vec{\text r}_0 + \vec{\text u}\text t + \frac{1}{2}\vec{\text a}\text t^2$. Substituting at t = 2 s :

$$\begin{aligned}\vec{\text r} = (2\ \hat{\text i} + 4\ \hat{\text j}) + (5\ \hat{\text i} + 4\ \hat{\text j})(2) + \frac{1}{2}(4\ \hat{\text i} + 4\ \hat{\text j})(2)^2 \\[1em] = (2\ \hat{\text i} + 4\ \hat{\text j}) + (10\ \hat{\text i} + 8\ \hat{\text j}) + (8\ \hat{\text i} + 8\ \hat{\text j}) \\[1em] = 20\ \hat{\text i} + 20\ \hat{\text j}\end{aligned}$$

The distance from the origin is the magnitude of this vector :

$|\vec{\text r}| = \sqrt{(20)^2 + (20)^2} = \sqrt{800} = 20\sqrt{2} \text { m}$

1 : 3 : 5 : 7

Reason — For a body dropped from rest, the distance in the n^th second is

$\text S_\text n = \text u + \frac{1}{2}\text g(2\text n - 1)$

With u = 0, putting n = 1, 2, 3, 4 in turn :

$\text S_1 = \frac{\text g}{2}, \quad \text S_2 = \frac{3\text g}{2}, \quad \text S_3 = \frac{5\text g}{2}, \quad \text S_4 = \frac{7\text g}{2}$

Dropping the common factor g/2, the ratio is

$\text S_1 : \text S_2 : \text S_3 : \text S_4 = 1 : 3 : 5 : 7$

1 : $\sqrt{3}$

Reason — On a displacement-time graph the velocity equals the slope, i.e. the tangent of the angle with the axis :

$\text v_1 = \tan 30°, \qquad \text v_2 = \tan 45°$

Taking the ratio :

$\frac{\text v_1}{\text v_2} = \frac{\tan 30°}{\tan 45°} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$

So the velocities are in the ratio 1 : $\sqrt{3}$.

uniformly accelerated having motion along a parabolic path.

Reason — The coordinates are x = 2 + 4t and y = 3t + 4t².

Along x :

$\frac{\text {dx}}{\text {dt}} = 4, \qquad \frac{\text d^2\text x}{\text {dt}^2} = 0$

Along y :

$\frac{\text {dy}}{\text {dt}} = 3 + 8\text t, \qquad \frac{\text d^2\text y}{\text {dt}^2} = 8$

There is acceleration only along y, and it is constant (independent of time), while x has none — so the motion is uniformly accelerated.

To find the path, express t from x = 2 + 4t as $\text t = \frac{\text x - 2}{4}$ and put it into y :

$\text y = 3\left(\frac{\text x - 2}{4}\right) + 4\left(\frac{\text x - 2}{4}\right)^2$

This has the form y = αx + βx², the equation of a parabola.

Hence the motion is uniformly accelerated along a parabolic path.

Reason — On a v-t plot the acceleration at any moment is the slope, so we read the slope over each stretch and match it to the a-t graph.

• O to A : the velocity is constant, so the acceleration is zero — the segment OP of the a-t graph.
• A to B : the velocity rises at a steady rate, so the acceleration is constant and positive — the segment QR.
• B to C : the velocity is constant again (uniform motion, v ≠ f(t)), so the acceleration is zero — the segment ST.
• C to D : the velocity falls at a steady rate, so there is a constant deceleration — the portion UV.
• At D the body comes to rest, marked by W on the a-t graph.

The a-t graph carrying these features (OP, QR, ST, UV and W) is graph (c).

$-\frac{2}{(2\text x + 1)^3}$

Reason — Starting from the given relation and differentiating with respect to x to reach the velocity :

$$\begin{aligned}\text t = \text x^2 + \text x \\[1em] \frac{\text {dt}}{\text {dx}} = 2\text x + 1 \\[1em] \Rightarrow 1 = (2\text x + 1)\frac{\text {dx}}{\text {dt}} \\[1em] \Rightarrow \text v = \frac{\text {dx}}{\text {dt}} = (2\text x + 1)^{-1}\end{aligned}$$

Now differentiate the velocity to get the acceleration, using the chain rule :

$$\begin{aligned}\text a = \frac{\text {dv}}{\text {dt}} = -(2\text x + 1)^{-2} \times 2 \times \frac{\text {dx}}{\text {dt}} \\[1em] = -2(2\text x + 1)^{-2} \times (2\text x + 1)^{-1} \\[1em] = -2(2\text x + 1)^{-3} \\[1em] = -\frac{2}{(2\text x + 1)^3}\end{aligned}$$

Note: The question incorrectly gives the relation as t² = x² + x, while the textbook’s solution is based on t = x² + x. If the printed relation is used, the result is $\frac{2}{(2x + 1)^3}$, which does not match any of the given options. Therefore, t² appears to be a printing error, and the intended relation t = x² + x has been used in the solution.

15 min, 120 km/h

Reason —

Given,

• Speed of the scooty, v_s = 60 km h^-1
• Same-direction interval, t₁ = 30 min = $\frac{1}{2}$ h
• Opposite-direction interval, t₂ = 10 min = $\frac{1}{6}$ h

Let the bus speed be v_B. The distance a following bus covers to catch the scooty is (v_B - v_s)t₁, and the distance an oncoming bus covers to reach her is (v_B + v_s)t₂. Since the buses leave at the same interval, both these distances are equal.

$$\begin{aligned}(\text v_\text B - \text v_\text s)\text t_1 = (\text v_\text B + \text v_\text s)\text t_2 \\[1em] (\text v_\text B - 60) \times \frac{1}{2} = (\text v_\text B + 60) \times \frac{1}{6} \\[1em] 3(\text v_\text B - 60) = \text v_\text B + 60 \\[1em] 3\text v_\text B - 180 = \text v_\text B + 60 \\[1em] 2\text v_\text B = 240 \\[1em] \Rightarrow \text v_\text B = 120 \text { km h}^{-1}\end{aligned}$$

The spacing between two consecutive buses is

$\text d = (\text v_\text B - \text v_\text s)\text t_1 = (120 - 60) \times \frac{1}{2} = 30 \text { km}$

So the service interval is

$\text T = \frac{\text d}{\text v_\text B} = \frac{30}{120} \text { h} = \frac{1}{4} \text { h} = 15 \text { min}$

Hence T = 15 min and the buses run at 120 km h^-1.

The value of α is 673.

Reason —

Given,

• Initial velocity components, u_x = 5 m s^-1, u_y = 0
• Acceleration components, a_x = 3 m s^-2, a_y = 2 m s^-2
• x-coordinate at the instant, s_x = 84 m

Velocity along x. Using v_x² = u_x² + 2a_xs_x :

$$\begin{aligned}\text v_\text x^2 = (5)^2 + 2 \times 3 \times 84 \\[1em] = 25 + 504 = 529 \\[1em] \Rightarrow \text v_\text x = 23 \text { m s}^{-1}\end{aligned}$$

Time to this instant. From v_x = u_x + a_xt :

$$\begin{aligned}23 = 5 + 3\ \text t \\[1em] \Rightarrow \text t = 6 \text { s}\end{aligned}$$

Velocity along y. From v_y = u_y + a_yt :

$\text v_\text y = 0 + (2 \times 6) = 12 \text { m s}^{-1}$

Resultant speed. Combining the two components :

$$\begin{aligned}\text v = \sqrt{\text v_\text x^2 + \text v_\text y^2} \\[1em] = \sqrt{(23)^2 + (12)^2} \\[1em] = \sqrt{529 + 144} \\[1em] = \sqrt{673} \text { m s}^{-1}\end{aligned}$$

Comparing with $\sqrt{\alpha}$, we get α = 673.

v₂ = 10 m/s

Reason —

Given,

• Velocity over the first stretch, v₁ = 5 m s^-1, covering a distance x
• Velocity over the next stretch, v₂, covering a distance $\frac{3}{2}$x
• Average velocity for the whole run = $\frac{50}{7}$ m s^-1

The total distance is

$\text x + \frac{3}{2}\text x = \frac{5}{2}\text x$

and the total time is the sum of the two leg-times :

$\frac{\text x}{\text v_1} + \frac{3\text x / 2}{\text v_2} = \frac{\text x}{5} + \frac{3\text x}{2\text v_2}$

Average velocity is total distance over total time :

$\frac{50}{7} = \frac{\frac{5}{2}\text x}{\frac{\text x}{5} + \frac{3\text x}{2\text v_2}}$

Cancelling x throughout :

$\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2\text v_2}}$

Rearranging for the bracket :

$\frac{1}{5} + \frac{3}{2\text v_2} = \frac{5}{2} \times \frac{7}{50} = \frac{7}{20}$

$$\begin{aligned}\frac{3}{2\text v_2} = \frac{7}{20} - \frac{4}{20} = \frac{3}{20} \\[1em] \Rightarrow 2\text v_2 = 20 \\[1em] \Rightarrow \text v_2 = 10 \text { m s}^{-1}\end{aligned}$$

Hence v₂ = 10 m s^-1.