Speed, Distance & Time

1 km = 1000 m and 1 h = 3600 s

$$\begin{aligned}3.6 \text{ km/h } = \frac{3.6 \times 1000 \text{ m}}{3600 \text{ s}}\\[1em] = \frac{3600}{3600} \text{ m/s }\\[1em] = 1 \text{ m/s }\end{aligned}$$

Hence, 3.6 km/h = 1 m/s.

1 km = 1000 m = 1000 × 100 cm = 1,00,000 cm and 1 h = 3600 s

$$\begin{aligned}54 \text{ km/h } = \frac{54 \times 1,00,000 \text{ cm}}{3600 \text{ s}}\\[1em] = \frac{54,00,000}{3600} \text{ cm/s }\\[1em] = 1500 \text{ cm/s }\end{aligned}$$

Hence, 54 km/h = 1500 cm/s.

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[ 1em] 60 \text{ m/s }= \frac{60 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] = \frac{60}{1000} \times 3600 \text{ km/h }\\[1em] = \frac{2,16,000}{1000} \text{ km/h }\\[1em] = 216 \text{ km/h }\end{aligned}$$

Hence, 60 m/s = 216 km/h.

$$\begin{aligned}1 \text{ cm } = \frac{1}{1,00,000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] 540 \text{ cm/s } = \frac{540 \times \frac{1}{1,00,000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] = \frac{540}{1,00,000} \times 3600 \text{ km/h }\\[1em] = \frac{19,44,000}{1,00,000} \text{ km/h }\\[1em] = 19.44 \text{ km/h }\end{aligned}$$

Hence, 540 cm/s = 19.44 km/h.

(i) Distance = 129 m and time = 3 min

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{129 \text{ m}}{3 \text{ min}} = 43 \text{ m/min }$

Hence, speed = 43 m/min.

(ii) Speed = 40 km/h and time = 45 min

$45 \text{ min } = \frac{45}{60} \text{ h } = \frac{3}{4} \text{ h }$

Distance = Speed × Time = $40 \times \frac{3}{4} \text{ km } = 30 \text{ km }$

Hence, distance = 30 km.

(iii) Distance = 720 m and speed = 48 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$$\begin{aligned}48 \text{ km/h } = \frac{48 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{48000}{3600} \text{ m/s } = \frac{40}{3} \text{ m/s }\\[1em] \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{720 \text{ m}}{\frac{40}{3} \text{ m/s}}\\[1em] = 720 \times \frac{3}{40} \text{ s }\\[1em] = 54 \text{ s }\end{aligned}$$

Hence, time = 54 sec.

	Distance	Speed	Time
(i)	129 m	43 m/min	3 min
(ii)	30 km	40 km/h	45 min
(iii)	720 m	48 km/h	54 sec

Extra distance to be covered = 400 m − 140 m = 260 m

Speed = 75 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$$\begin{aligned}75 \text{ km/h } = \frac{75 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{75000}{3600} \text{ m/s } = \frac{125}{6} \text{ m/s }\\[1em] \text{Extra time} = \frac{\text{Extra distance}}{\text{Speed}} = \frac{260 \text{ m}}{\frac{125}{6} \text{ m/s}}\\[1em] = 260 \times \frac{6}{125} \text{ s }\\[1em] = \frac{1560}{125} \text{ s }\\[1em] = 12.48 \text{ s }\end{aligned}$$

Hence, extra time taken = 12.48 sec.

To compare the two speeds, both must be in the same unit.

Converting 72 km/h into m/s :

1 km = 1000 m and 1 h = 3600 s

$72 \text{ km/h } = \frac{72 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{72000}{3600} \text{ m/s } = 20 \text{ m/s }$

Since, 20 m/s < 21 m/s

⇒ 72 km/h < 21 m/s

Hence, 21 m/s is greater.

Distance of the journey = Speed × Time = 90 km/h × 3 h = 270 km

New time = 5 h

New speed = $\frac{\text{Distance}}{\text{Time}} = \frac{270 \text{ km}}{5 \text{ h}} = 54 \text{ km/h }$

Hence, the required speed = 54 km/h.

Speed = 60 km/h

Time = 50 min = $\frac{50}{60} \text{ h } = \frac{5}{6} \text{ h }$

Distance = Speed × Time = $60 \times \frac{5}{6} \text{ km } = 50 \text{ km }$

When the speed is halved,

$$\begin{aligned}\text{New speed} = \frac{60}{2} \text{ km/h } = 30 \text{ km/h }\\[1em] \text{New time} = \frac{50 \text{ km}}{30 \text{ km/h}}\\[1em] = \frac{5}{3} \text{ h }\\[1em] = \frac{5}{3} \times 60 \text{ min }\\[1em] = 100 \text{ min }\end{aligned}$$

Extra time = 100 min − 50 min = 50 min

Hence, 50 minutes extra time will be needed.

Speed = $\frac{\text{Distance}}{\text{Time}} = \frac{800 \text{ m}}{5 \text{ min}} = 160 \text{ m/min }$

(i) Time = half an hour = 30 min

Distance = Speed × Time = 160 × 30 m = 4800 m = 4.8 km

Hence, she will cover 4800 m = 4.8 km.

(ii) Distance = 6.4 km = 6.4 × 1000 m = 6400 m

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{6400 \text{ m}}{160 \text{ m/min}} = 40 \text{ min }$

Hence, she will take 40 minutes.

Distance from Delhi to Agra = Speed × Time = 60 km/h × 4 h = 240 km

(i) Increased speed = 60 km/h + 20 km/h = 80 km/h

Time = $\frac{240 \text{ km}}{80 \text{ km/h}} = 3 \text{ h }$

Hence, the bus will take 3 hours.

(ii) Reduced speed = 60 km/h − 20 km/h = 40 km/h

Time = $\frac{240 \text{ km}}{40 \text{ km/h}} = 6 \text{ h }$

Extra time = 6 h − 4 h = 2 h

Hence, the bus will take 2 hours extra.

(i) Speed = 6 km/h and time = 8 min

$$\begin{aligned}8 \text{ min } = \frac{8}{60} \text{ h } = \frac{2}{15} \text{ h }\\[1em] \text{Distance} = \text{Speed} \times \text{Time} = 6 \times \frac{2}{15} \text{ km }\\[1em] = \frac{12}{15} \text{ km }\\[1em] = 0.8 \text{ km } = 800 \text{ m }\end{aligned}$$

Hence, it will walk 0.8 km = 800 m.

(ii) Distance = 600 m = $\frac{600}{1000} \text{ km } = 0.6 \text{ km }$

$$\begin{aligned}\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{0.6 \text{ km}}{6 \text{ km/h}}\\[1em] = 0.1 \text{ h }\\[1em] = 0.1 \times 60 \text{ min } = 6 \text{ min }\end{aligned}$$

Hence, it will take 6 minutes.

Total distance covered = 80 km + 80 km = 160 km

Total time taken = 5 h

Speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{160 \text{ km}}{5 \text{ h}} = 32 \text{ km/h }$

Hence, the speed of Manoj = 32 km/h.

Let car A meet car B after x hours.

Car A starts at 6 a.m., so car A moves for x hours.

Car B starts at 7 a.m., i.e. one hour later, so car B moves for (x − 1) hours.

Distance moved by car A = 40 × x = 40x km

Distance moved by car B = 50 × (x − 1) = 50(x − 1) km

When the cars meet,

Distance moved by car A + Distance moved by car B = 220 km

⇒ 40x + 50(x - 1) = 220

⇒ 40x + 50x - 50 = 220

⇒ 90x = 270

$\Rightarrow x = \frac{270}{90}$

⇒ x = 3

So, the cars meet 3 hours after 6 a.m.

Meeting time = (6 + 3) a.m. = 9 a.m.

Hence, cars A and B will meet at 9 a.m.

Speed = 45 km/h

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$45 \text{ km/h } = \frac{45 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{45000}{3600} \text{ m/s } = 12.5 \text{ m/s }$

Time = 24 s

Distance = Speed × Time = 12.5 × 24 m = 300 m

Hence, the distance covered = 300 m.

Converting the speed of the first car into m/s :

1 km = 1000 m and 1 h = 3600 s

$72 \text{ km/h } = \frac{72 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{72000}{3600} \text{ m/s } = 20 \text{ m/s }$

Speed of the second car = 16 m/s

As both the cars move in the same direction,

Relative speed = 20 m/s − 16 m/s = 4 m/s

(i) Difference of distances in 1 second = 4 m/s × 1 s = 4 m

Hence, the difference of distances in 1 second = 4 m.

(ii) Difference of distances in 15 seconds = 4 m/s × 15 s = 60 m

Hence, the difference of distances in 15 seconds = 60 m.

Original speed = $\frac{\text{Distance}}{\text{Time}} = \frac{6 \text{ km}}{8 \text{ min}} = 0.75 \text{ km/min }$

New speed = $\frac{6 \text{ km}}{4 \text{ min}} = 1.5 \text{ km/min }$

Increase in speed = 1.5 km/min − 0.75 km/min = 0.75 km/min

Hence, the speed must be increased by 0.75 km/min.

Total distance covered = 200 km + 126 km + 143 km = 469 km

Total time taken = 2 h + 2 h + 3 h = 7 h

$$\begin{aligned}\text{Average speed } = \frac{\text{Total distance covered}}{\text{Total time taken}}\\[1em] = \frac{469 \text{ km}}{7 \text{ h}}\\[1em] = 67 \text{ km/h }\end{aligned}$$

Hence, the average speed of the train = 67 km/h.

Distance covered in the first part = 72 km/h × 5 h = 360 km

Distance covered in the second part = 90 km/h × 4 h = 360 km

Total distance covered = 360 km + 360 km = 720 km

Total time taken = 5 h + 4 h = 9 h

Average speed = $\frac{720 \text{ km}}{9 \text{ h}} = 80 \text{ km/h }$

Hence, the average speed of the bus = 80 km/h.

For the first part :

Distance = 60 km and speed = 40 km/h

$\text{Time taken} = \frac{60 \text{ km}}{40 \text{ km/h}} = \frac{3}{2} \text{ h }$

For the remaining part :

Distance = 80 km − 60 km = 20 km and speed = 20 km/h

Time taken = $\frac{20 \text{ km}}{20 \text{ km/h}} = 1 \text{ h }$

Total time taken = $\frac{3}{2} \text{ h } + 1 \text{ h } = \frac{5}{2} \text{ h }$

$$\begin{aligned}\text{Average speed} = \frac{80 \text{ km}}{\frac{5}{2} \text{ h}}\\[1em] = 80 \times \frac{2}{5} \text{ km/h }\\[1em] = 32 \text{ km/h }\end{aligned}$$

Hence, the average speed = 32 km h^-1.

Let the distance between stations P and Q be x km.

Time taken from P to Q = $\frac{x}{60} \text{ h }$

Time taken from Q to P = $\frac{x}{30} \text{ h }$

Total distance covered = x + x = 2x km

$$\begin{aligned}\text{Total time taken} = \frac{x}{60} + \frac{x}{30}\\[1em] = \frac{x + 2x}{60}\\[1em] = \frac{3x}{60} = \frac{x}{20} \text{ h }\end{aligned}$$

Calculating average speed,

$$\begin{aligned}\text{Average speed } = \frac{\text{Total distance covered}}{\text{Total time taken}}\\[1em] = \frac{2x}{\frac{x}{20}}\\[1em] = 2x \times \frac{20}{x}\\[1em] = 40 \text{ km/h }\end{aligned}$$

Hence, the average speed for the entire journey = 40 km h^-1.

(i) Distance covered at 75 km/h = 300 km − (85 km + 90 km)

= 300 km − 175 km = 125 km

Hence, the distance covered at a speed of 75 km/h = 125 km.

(ii) Time taken for the first part = $\frac{85 \text{ km}}{51 \text{ km/h}} = \frac{5}{3} \text{ h }$

Time taken for the second part = $\frac{90 \text{ km}}{135 \text{ km/h}} = \frac{2}{3} \text{ h }$

Time taken for the third part = $\frac{125 \text{ km}}{75 \text{ km/h}} = \frac{5}{3} \text{ h }$

$$\begin{aligned}\text{Total time taken} = \frac{5}{3} + \frac{2}{3} + \frac{5}{3}\\[1em] = \frac{12}{3} \text{ h }\\[1em] = 4 \text{ h }\end{aligned}$$

Hence, the total time taken = 4 hours.

(iii) Average speed = $\frac{300 \text{ km}}{4 \text{ h}} = 75 \text{ km/h }$

Hence, the average speed for the whole journey = 75 km h^-1.

Time taken for the first part = $\frac{72 \text{ km}}{36 \text{ km/h}} = 2 \text{ h }$

Time taken for the second part = $\frac{135 \text{ km}}{45 \text{ km/h}} = 3 \text{ h }$

Total distance covered = 72 km + 135 km = 207 km

Total time taken = 2 h + 3 h = 5 h

Average speed = $\frac{207 \text{ km}}{5 \text{ h}} = 41.4 \text{ km/h }$

Hence, the average speed of the motorcycle = 41.4 km h^-1.

(i) When two bodies move in opposite directions, their relative speed is the sum of their speeds.

Relative speed = 120 km/h + 75 km/h = 195 km/h

Hence, their relative speed = 195 km h^-1.

(ii) When two bodies move in the same direction, their relative speed is the difference of their speeds.

Relative speed = 120 km/h − 75 km/h = 45 km/h

Hence, their relative speed = 45 km h^-1.

In crossing a pole, the train covers a distance equal to its own length.

Distance = 900 m and time = 45 s

Speed = $\frac{900 \text{ m}}{45 \text{ s}} = 20 \text{ m/s }$

Converting this speed into km/h :

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] 20 \text{ m/s } = \frac{20 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] = \frac{20}{1000} \times 3600 \text{ km/h }\\[1em] = 72 \text{ km/h }\end{aligned}$$

Hence, the speed of the train = 72 km h^-1.

In passing a standing man, the train covers a distance equal to its own length.

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$90 \text{ km/h } = \frac{90 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{90000}{3600} \text{ m/s } = 25 \text{ m/s }$

Time = 8 s

Length of the train = Speed × Time = 25 × 8 m = 200 m

Hence, the length of the train = 200 m.

In passing a platform, the train covers a distance equal to the sum of its own length and the length of the platform.

Distance covered = 100 m + 200 m = 300 m

Time = 20 s

Speed = $\frac{300 \text{ m}}{20 \text{ s}} = 15 \text{ m/s }$

Converting this speed into km/h :

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] 15 \text{ m/s } = \frac{15 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] = \frac{15}{1000} \times 3600 \text{ km/h }\\[1em] = 54 \text{ km/h }\end{aligned}$$

Hence, the speed of the train = 15 ms^-1 = 54 km h^-1.

(i) Relative speed (same direction) = 80 km/h − 50 km/h = 30 km/h

Distance between them in 3 h = 30 km/h × 3 h = 90 km

Hence, the distance between the two cars = 90 km.

(ii) Relative speed (opposite directions) = 80 km/h + 50 km/h = 130 km/h

Distance between them in 3 h = 130 km/h × 3 h = 390 km

Hence, the distance between the two cars = 390 km.

In passing a platform, the train covers a distance equal to the sum of its own length and the length of the platform.

Distance covered = Length of train + Length of platform

= 80 m + 220 m = 300 m

Converting the speed into m/s :

1 km = 1000 m and 1 h = 3600 s

$45 \text{ km/h } = \frac{45 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{45000}{3600} \text{ m/s } = 12.5 \text{ m/s }$

Time = $\frac{\text{Distance}}{\text{Speed}} = \frac{300 \text{ m}}{12.5 \text{ m/s}} = 24 \text{ s }$

Hence, the time taken by the train = 24 sec.

Converting both the speeds into m/s :

1 km = 1000 m and 1 h = 3600 s

Speed of the bus = $90 \text{ km/h } = \frac{90 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{90000}{3600} \text{ m/s } = 25 \text{ m/s }$

Speed of the truck = $72 \text{ km/h } = \frac{72 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{72000}{3600} \text{ m/s } = 20 \text{ m/s }$

(i) Relative speed (same direction) = 25 m/s − 20 m/s = 5 m/s

Distance between them in 20 s = 5 m/s × 20 s = 100 m

Hence, the distance between the two = 100 m.

(ii) Relative speed (opposite directions) = 25 m/s + 20 m/s = 45 m/s

Distance between them in 20 s = 45 m/s × 20 s = 900 m

Hence, the distance between the two = 900 m.

Let the length of the train be $x$ m and its speed be $v$ m/s.

While passing the pole, the train covers a distance equal to its own length.

$x = v \times 2 \qquad \text{...(1)}$

While passing the platform, the train covers a distance equal to the sum of its own length and the length of the platform.

$x + 50 = v \times \frac{9}{2} \qquad \text{...(2)}$

Subtracting equation (1) from equation (2) :

$$\begin{aligned}\Rightarrow (x + 50) - x = \frac{9}{2}v - 2v\\[1em] \Rightarrow 50 = \frac{9v - 4v}{2}\\[1em] \Rightarrow 50 = \frac{5v}{2}\\[1em] \Rightarrow v = \frac{50 \times 2}{5}\\[1em] \Rightarrow v = 20 \text{ m/s }\end{aligned}$$

Substituting $v = 20$ m/s in equation (1) :

$x = 20 \times 2 = 40 \text{ m }$

Converting the speed into km/h :

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] \Rightarrow 20 \text{ m/s } = \frac{20 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] \Rightarrow 20 \text{ m/s } = \frac{20}{1000} \times 3600 \text{ km/h }\\[1em] \Rightarrow 20 \text{ m/s }= 72 \text{ km/h }\end{aligned}$$

Hence, the length of the train = 40 m and its speed = 72 km h^-1.

Let the length of the train be $x$ m and its speed be $v$ m/s.

While passing the man, the train covers a distance equal to its own length.

$x = v \times 6 \qquad \text{...(1)}$

While passing the platform, the train covers a distance equal to the sum of its own length and the length of the platform.

$x + 225 = v \times 21 \qquad \text{...(2)}$

Subtracting equation (1) from equation (2) :

$$\begin{aligned}\Rightarrow (x + 225) - x = 21v - 6v\\[1em] \Rightarrow 225 = 15v\\[1em] \Rightarrow v = \frac{225}{15}\\[1em] \Rightarrow v = 15 \text{ m/s }\end{aligned}$$

Speed of the train = 15 m/s

(i) Substituting $v = 15$ m/s in equation (1) :

$x = 15 \times 6 = 90 \text{ m }$

Hence, the length of the train = 90 m.

(ii) Converting the speed into km/h :

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] \Rightarrow 15 \text{ m/s } = \frac{15 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] \Rightarrow 15 \text{ m/s } = \frac{15}{1000} \times 3600 \text{ km/h }\\[1em] \Rightarrow 15 \text{ m/s } = 54 \text{ km/h }\end{aligned}$$

Hence, the speed of the train = 54 km h^-1.

60 min = 1 hour

⇒ $1 \text{ min } = \frac{1}{60} \text{ h }$

1000 m = 1 km

$$\begin{aligned}\Rightarrow 90 \text{ m } = \Big(\frac{1}{1000} \times 90\Big) \text{ km } = 0.09 \text{ km }\\[1em] \text{Speed} = \frac{0.09 \text{ km}}{\frac{1}{60} \text{ h}}\\[1em] = 0.09 \times 60 \text{ km/h }\\[1em] = 5.4 \text{ km/h }\end{aligned}$$

Hence, option 2 is the correct option.

Time taken for the first 100 km = $\frac{100}{20} = 5 \text{ h }$

Time taken for the next 100 km = $\frac{100}{30} = \frac{10}{3} \text{ h }$

Total distance = 100 km + 100 km = 200 km

$$\begin{aligned}\text{Total time} = 5 + \frac{10}{3}\\[1em] = \frac{15 + 10}{3}\\[1em] = \frac{25}{3} \text{ h }\\[1em] \text{Average speed} = \frac{200}{\frac{25}{3}}\\[1em] = 200 \times \frac{3}{25}\\[1em] = 24 \text{ km/h }\end{aligned}$$

As 24 km/h is not among the options given,

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\text{Distance} = 100 \text{ m } = \frac{100}{1000} \text{ km } = 0.1 \text{ km }\\[1em]\text{Time} = 24 \text{ s } = \frac{24}{3600} \text{ h } = \frac{1}{150} \text{ h }\\[1em] \text{Speed} = \frac{0.1 \text{ km}}{\frac{1}{150} \text{ h}}\\[1em] = 0.1 \times 150 \text{ km/h }\\[1em] = 15 \text{ km/h }\end{aligned}$$

Hence, option 3 is the correct option.

Total distance covered = 200 m + 400 m = 600 m

Total time taken = 20 min + 30 min = 50 min

Average speed = $\frac{600 \text{ m}}{50 \text{ min}} = 12 \text{ m/min }$

Hence, option 2 is the correct option.

Time taken for the first 300 m = $\frac{300}{20} = 15 \text{ min }$

Time taken for the next 300 m = $\frac{300}{25} = 12 \text{ min }$

Total time taken = 15 min + 12 min = 27 min

Hence, option 4 is the correct option.

Time taken for the first 300 km = $\frac{300}{20} = 15 \text{ h }$

Time taken for the next 300 km = $\frac{300}{30} = 10 \text{ h }$

Total distance = 300 km + 300 km = 600 km

Total time = 15 h + 10 h = 25 h

Average speed = $\frac{600 \text{ km}}{25 \text{ h}} = 24 \text{ km/h }$

Hence, option 3 is the correct option.

As they move in opposite directions,

Relative speed = 18 m/s + 22 m/s = 40 m/s

Distance between them in 10 s = 40 × 10 m = 400 m

Hence, option 4 is the correct option.

As they move in the same direction,

Relative speed = 35 m/s − 25 m/s = 10 m/s

Distance between them in 20 s = 10 × 20 m = 200 m

Hence, option 1 is the correct option.

Average speed = $\frac{\text{Total distance covered}}{\text{Total time taken}}$

The average speed cannot be found by taking the average of the two speeds, as neither the distances covered nor the times taken at these speeds are given.

Hence, option 4 is the correct option.

Time = 45 min = 45 × 60 s = 2700 s

Distance = Speed × Time = 20 × 2700 m = 54000 m

New time = 30 min = 30 × 60 s = 1800 s

x = $\frac{54000 \text{ m}}{1800 \text{ s}} = 30 \text{ m/s }$

Hence, option 3 is the correct option.

Remaining distance = 500 m − 200 m = 300 m

Remaining time = 40 s − 20 s = 20 s

Average speed for the remaining distance = $\frac{300 \text{ m}}{20 \text{ s}} = 15 \text{ m/s }$

Hence, option 3 is the correct option.

Distance = Speed × Time = 50 × 4 km = 200 km

Reduced speed = 50 km/h − 10 km/h = 40 km/h

New time = $\frac{200 \text{ km}}{40 \text{ km/h}} = 5 \text{ h }$

Increase in time = 5 h − 4 h = 1 h

Hence, option 4 is the correct option.

Total distance covered = 200 km + 680 km = 880 km

Total time taken = 2 h + 3 h = 5 h

Average speed = $\frac{880 \text{ km}}{5 \text{ h}} = 176 \text{ km/h }$

Hence, option 2 is the correct option.

As the cars move in opposite directions,

Relative speed = 40 km/h + 50 km/h = 90 km/h

Distance between them in 3 h = 90 × 3 km = 270 km

Hence, option 4 is the correct option.

As the cars move in the same direction,

Relative speed = 50 km/h − 40 km/h = 10 km/h

Distance between them in 3 h = 10 × 3 km = 30 km

Hence, option 1 is the correct option.

Converting 1 m/s into km/h,

$$\begin{aligned}1 \text{ m/s } = \frac{\frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] = \frac{3600}{1000} \text{ km/h }\\[1em] = 3.6 \text{ km/h }\end{aligned}$$

So, Statement 1 is true.

Converting 1 km/h into m/s,

$$\begin{aligned}1 \text{ km/h } = \frac{1000 \text{ m}}{3600 \text{ s}}\\[1em] = \frac{5}{18} \text{ m/s }\\[1em] \approx 0.28 \text{ m/s }\end{aligned}$$

So, Statement 2 is false.

Hence, option 3 is the correct option.

Given,

A cyclist covers 1.4 km in 240 seconds.

$$\begin{aligned}\text{Time} = 240 \text{ s } = \frac{240}{3600} \text{ h } = \frac{1}{15} \text{ h }\\[1em] \text{Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] \Rightarrow \text{Speed} = \frac{1.4 \text{ km}}{\frac{1}{15} \text{ h}}\\[1em] = 1.4 \times 15 \text{ km/h }\\[1em] = 21 \text{ km/h }\end{aligned}$$

So, Assertion (A) is true.

Speed is defined as the distance travelled by a moving body in unit time. So, Reason (R) is true.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] \Rightarrow 7.5 \text{ m/s } = \frac{7.5 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] \Rightarrow 7.5 \text{ m/s } = \frac{7.5}{1000} \times 3600 \text{ km/h }\\[1em] \Rightarrow 7.5 \text{ m/s } = 27 \text{ km/h }\end{aligned}$$

So, Assertion (A) is true.

$$\begin{aligned}1 \text{ m/s } = \frac{\frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] \Rightarrow 1 \text{ m/s } = \frac{3600}{1000} \text{ km/h }\\[1em] \Rightarrow 1 \text{ m/s } = \frac{18}{5} \text{ km/h }\end{aligned}$$

So, a speed in m/s is converted into km/h by multiplying it by $\frac{18}{5}$. So, Reason (R) is true.

Hence, option 3 is the correct option.

Speed = 12 km/h

Time = 35 min = $\frac{35}{60} \text{ h } = \frac{7}{12} \text{ h }$

Distance = Speed × Time = $12 \times \frac{7}{12} \text{ km } = 7 \text{ km }$

and not 7.5 km. So, Assertion (A) is false.

Since $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$, we get Distance covered = Speed × Time. So, Reason (R) is true.

Hence, option 2 is the correct option.

Solving,

1 min = 60 s

$$\begin{aligned}150 \text{ m/min } = \frac{150 \text{ m}}{60 \text{ s}} = 2.5 \text{ m/s }\\[1em] 1 \text{ m } = \frac{1}{1000} \text{ km and } 1 \text{ s } = \frac{1}{3600} \text{ h }\\[1em] \Rightarrow 2.5 \text{ m/s } = \frac{2.5 \times \frac{1}{1000} \text{ km}}{\frac{1}{3600} \text{ h}}\\[1em] \Rightarrow 2.5 \text{ m/s } = \frac{2.5}{1000} \times 3600 \text{ km/h }\\[1em] \Rightarrow 2.5 \text{ m/s } = 9 \text{ km/h }\end{aligned}$$

So, Assertion (A) is true.

Speed in cm per sec = $\frac{\text{Distance in centimeter (cm)}}{\text{Time in seconds (s)}}$, when the distance is in cm and the time is in seconds, the speed obtained is in cm per sec. So, Reason (R) is true.

Hence, option 3 is the correct option.