Speed, Distance & Time

(i) 90 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$:

$$\begin{aligned}\therefore 90 \text { km/hr} =\Big(90 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = (5 \times 5) \text{ m/sec} \\[1em] = 25 \text{ m/sec}\end{aligned}$$

Hence, the answer is 25 m/sec.

(ii) 27 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$:

$$\begin{aligned}\therefore 27 \text { km/hr} = \Big(27 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(3 \times \frac{5}{2}\Big)\text{ m/sec} \\[1em] = \frac{15}{2}\text{ m/sec} \\[1em] = 7.5\text{ m/sec}\end{aligned}$$

Hence, the answer is 7.5 m/sec.

(iii) 315 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$:

$$\begin{aligned}\therefore 315 \text { km/hr} = \Big(315 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \frac{1575}{18}\text{ m/sec} \\[1em] = 87.5\text{ m/sec}\end{aligned}$$

Hence, the answer is 87.5 m/sec.

(iv) 16.2 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$:

$$\begin{aligned}\therefore 16.2 \text { km/hr} = \Big(16.2 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \frac{81}{18}\text{ m/sec} \\[1em] = 4.5\text{ m/sec}\end{aligned}$$

Hence, the answer is 4.5 m/sec.

(i) 35 m/sec

To convert m/sec to km/hr, multiply by $\frac{18}{5}$:

$$\begin{aligned}\therefore 35 \text { m/sec} = \Big(35 \times \frac{18}{5}\Big)\text{ km/hr} \\[1em] = (7 \times 18) \text{ km/hr} \\[1em] = 126 \text{ km/hr}\end{aligned}$$

Hence, the answer is 126 km/hr.

(ii) 14 m/sec

To convert m/sec to km/hr, multiply by $\frac{18}{5}$:

$$\begin{aligned}\therefore 14 \text { m/sec} = \Big(14 \times \frac{18}{5}\Big)\text{ km/hr} \\[1em] = \frac{252}{5}\text{ km/hr} \\[1em] = 50.4\text{ km/hr}\end{aligned}$$

Hence, the answer is 50.4 km/hr.

(iii) 60 m/sec

To convert m/sec to km/hr, multiply by $\frac{18}{5}$:

$$\begin{aligned}\therefore 60 \text { m/sec} =\Big(60 \times \frac{18}{5}\Big)\text{ km/hr} \\[1em] = (12 \times 18) \text{ km/hr} \\[1em] = 216 \text{ km/hr}\end{aligned}$$

Hence, the answer is 216 km/hr.

(iv) 3.5 m/sec

To convert m/sec to km/hr, multiply by $\frac{18}{5}$:

$$\begin{aligned}\therefore 3.5 \text { m/sec} = \Big(3.5 \times \frac{18}{5}\Big)\text{ km/hr} \\[1em] = \frac{63}{5}\text{ km/hr} \\[1em] = 12.6\text{ km/hr}\end{aligned}$$

Hence, the answer is 12.6 km/hr.

Given:

Distance = 258 km

Time = 3 hours 35 mins

Let's convert Time to hours:

1 hour = 60 mins

∴ 3 hours 35 mins = $3\frac{35}{60} \text{ hr} = 3\frac{7}{12} \text{ hr} = \frac{43}{12} \text{ hr}$

And

$$\begin{aligned}\text {Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \left(\frac{258}{\frac{43}{12}}\right)\text{ km/hr} \\[1em] = \Big(258 \times \frac{12}{43}\Big)\text{ km/hr} \\[1em] = 6 \times 12\text{ km/hr} \\[1em] = 72 \text{ km/hr}\end{aligned}$$

The speed of the car is 72 km/hr.

Given:

Distance = 14 km

Time = 8 minutes

Let's convert Time to hours:

1 hour = 60 mins

∴ 8 minutes = $\frac{8}{60} \text{ hr} = \frac{2}{15} \text{ hr}$

And

$$\begin{aligned}\text {Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \left(\frac{14}{\frac{2}{15}}\right)\text{ km/hr} \\[1em] = \Big(14 \times \frac{15}{2}\Big)\text{ km/hr} \\[1em] = 7 \times 15\text{ km/hr} \\[1em] = 105 \text{ km/hr}\end{aligned}$$

The speed of the train is 105 km/hr.

Given:

Distance = 8 km

Time = 1 hour 15 minutes

Let's convert Time to hours

1 hour = 60 mins

∴ 1 hour 15 minutes = $1\frac{15}{60} \text{ hr} = 1\frac{1}{4} \text{ hr} = \frac{5}{4} \text{ hr}$

And

$$\begin{aligned}\text {Speed} = \frac{\text{Distance}}{\text{Time}} \\[1em] = \left(\frac{8}{\frac{5}{4}}\right)\text{ km/hr} \\[1em] = \Big(8 \times \frac{4}{5}\Big)\text{ km/hr} \\[1em] = \frac{32}{5}\text{ km/hr} \\[1em] = 6.4 \text{ km/hr}\end{aligned}$$

The speed of the man is 6.4 km/hr.

Given:

Speed = 12 km/hr

Time = 36 minutes

Let's convert Time to hours:

1 hour = 60 mins

∴ 36 minutes = $\frac{36}{60} \text{ hr} = \frac{3}{5} \text{ hr}$

And

Distance = Speed x Time

Substituting the values in above, we get:

$$\begin{aligned}\text {Distance} = \Big(12 \times \frac{3}{5}\Big)\text{ km} \\[1em] = \frac{36}{5}\text{ km} \\[1em] = 7.2 \text{ km}\end{aligned}$$

The distance covered by Rahul is 7.2 km.

Given:

Speed = 4 km/hr

Distance = 600 metres

Let's convert Distance to km:

1 km = 1000 m

∴ 600 m = $\frac{600}{1000} \text{ km} = 0.6 \text{ km}$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

Time = $\frac{0.6}{4}$ hours = 0.15 hours

Let's convert time to minutes:

1 hour = 60 minutes

∴ 0.15 hours = 0.15 x 60 minutes = 9 minutes

The time taken by Ravi is 9 minutes.

Given:

Speed = 45 km/hr

Let's convert speed into m/sec:

$$\begin{aligned}45 \text { km/hr} = \Big(45 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \frac{225}{18}\text{ m/sec} \\[1em] = 12.5\text{ m/sec}\end{aligned}$$

Time = 8 seconds

Distance = Speed x Time

Substituting the values in above, we get:

Distance = 12.5 x 8 m = 100 m

The distance covered by train is 100 m.

Let's convert 36 km/hr into m/sec:

$$\begin{aligned}36 \text { km/hr} = \Big(36 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = (2 \times 5)\text{ m/sec} \\[1em] = 10 \text{ m/sec}\end{aligned}$$

Now, compare 10 m/sec and 10.5 m/sec:

10.5 m/sec > 10 m/sec

10.5 m/sec is more.

Given:

Speed = 45 km/hr

Time = 6 hours

Distance = Speed x Time

Substituting the values in above, we get:

Distance = 45 x 6 km = 270 km

Now, let's find the required speed for the new time:

Distance = 270 km, New Time = 5 hours

New Speed = $\frac{\text{Distance}}{\text{New Time}}$

Substituting the values in above, we get:

New Speed = $\frac{270}{5}$ km/hr

New Speed = 54 km/hr

Speed of the car to complete the journey in 5 hours is 54 km/hr.

Given:

Speed = 45 km/hr

Time = 50 minutes

Let's convert time to hour:

1 hour = 60 minutes

∴ 50 minutes = $\frac{50}{60}\text{ hr} = \frac{5}{6}\text{ hr}$

Distance = Speed x Time

Substituting the values in above, we get:

Distance = $45 \times \frac{5}{6}$ km = $\frac{225}{6}$ km

Let's find the time for the new speed:

Distance = $\frac{225}{6}$ km, New Speed = 30 km/hr

Time = $\frac{\text{Distance}}{\text{New Speed}}$

Substituting the values in above, we get:

Time = $\frac{\frac{225}{6}}{30}\text{ hours}$

$$\begin{aligned}= \Big(\frac{225}{6} \times \frac{1}{30}\Big)\text{ hours} \\[1em] = \Big(\frac{45}{6} \times \frac{1}{6}\Big)\text{ hours} \\[1em] = \frac{45}{36} \text{ hours} \\[1em] = 1.25 \text{ hours}\end{aligned}$$

1.25 hours = 1 hour + (0.25 x 60) minutes = 1 hour 15 minutes

Time taken by the car is 1 hour 15 minutes.

Given:

Distance : (D1) = 100 km, (D2) = 64 km, (D3) = 72 km

Time : (T1) = 2 hrs, (T2) = 1 hr, (T3) = 2 hrs

Total Distance = 100 km + 64 km + 72 km = 236 km

Total Time = 2 hrs + 1 hrs + 2 hrs = 5 hrs

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{236}{5}$ km/hr = 47.2 km/hr

Average speed of the bus is 47.2 km/hr.

Given:

Distances: (D1) = 2500 km, D2 = 1200 km, D3 = 500 km

Speed: (S1) = 500 km/hr, (S2) = 400 km/hr, (S3) = 250 km/hr

Let's find time taken for each part:

Time = $\frac{\text{Distance}}{\text{Speed}}$

T1 = $\frac{D1}{S1} = \frac{2500}{500}$ = 5 hrs

T2 = $\frac{D2}{S2} = \frac{1200}{400}$ = 3 hrs

T3 = $\frac{D3}{S3} = \frac{500}{250}$ = 2 hrs

Total Distance = 2500 km + 1200 km + 500 km = 4200 km

Total Time = 5 hrs + 3 hrs + 2 hrs = 10 hrs

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{4200}{10}$ km/hr = 420 km/hr

Average speed of the aeroplane is 420 km/hr.

Given:

Speed: S1 = 36 km/hr, S2 = 45 km/hr

Time: T1 = 5 hrs, T2 = 4 hrs

Let's find distance covered in each part:

Distance = Speed x Time

D1 = S1 x T1 = 36 x 5 = 180 km

D2 = S2 x T2 = 45 x 4 = 180 km

Total Distance = 180 km + 180 km = 360 km

Total Time = 5 hrs + 4 hrs = 9 hrs

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{360}{9}$ km/hr = 40 km/hr

Average speed of the scooterist is 40 km/hr.

Given:

Distance (one way) = 4 km

Total Distance = 4 km (to school) + 4 km (back home) = 8 km

Total Time = $2\frac{1}{2} \text{ hours} = 2.5 \text{ hours}$

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{8}{2.5}$ km/hr = 3.2 km/hr

Average speed of Raju is 3.2 km/hr.

Given:

Distance = Length of train = 150 m

Speed = 60 km/hr

Let's convert speed to m/sec:

$$\begin{aligned}60 \text { km/hr} = \Big(60 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(10 \times \frac{5}{3}\Big)\text{ m/sec} \\[1em] = \frac{50}{3} \text{ m/sec}\end{aligned}$$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

$$\begin{aligned}\text {Time} = \frac{150}{\frac{50}{3}} \text{ seconds} \\[1em] = 150 \times \frac{3}{50}\text{ seconds} \\[1em] = 3 \times \frac{3}{1}\text{ seconds} \\[1em] = 3 \times 3\text{ seconds} \\[1em] = 9 \text{ seconds}\end{aligned}$$

Hence, the train will cross a man in 9 seconds.

Given:

Distance = Length of train = 175 m

Speed = 63 km/hr

Let's convert speed to m/sec:

$$\begin{aligned}63 \text { km/hr} = \Big(63 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(7 \times \frac{5}{2}\Big)\text{ m/sec} \\[1em] = \frac{35}{2} \text{ m/sec}\end{aligned}$$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

$$\begin{aligned}\text {Time} = \frac{175}{\frac{35}{2}} \\[1em] = 175 \times \frac{2}{35}\text{ seconds} \\[1em] = 5 \times \frac{2}{1}\text{ seconds} \\[1em] = 5 \times 2\text{ seconds} \\[1em] = 10 \text{ seconds}\end{aligned}$$

Hence, the train will cross an electric pole in 10 seconds.

Given:

Distance = Length of train = 150 m

Time = 6 seconds

Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting the values in above, we get:

Speed = $\frac{150}{6}$ m/sec = 25 m/sec

Let's convert speed to km/hr:

$$\begin{aligned}\therefore 25 \text{ m/sec} = 25 \times \frac{18}{5} \text{ km/hr} \\[1em] = 5 \times \frac{18}{1} \text{ km/hr} \\[1em] = 5 \times 18 \text{ km/hr} \\[1em] = 90 \text{ km/hr}\end{aligned}$$

Hence, speed of the train is 90 km/hr.

Given:

Distance (D1) = Length of train = 120 m

Distance (D2) = Length of bridge = 180 m

Total Distance = 120 m + 180 m = 300 m

Speed = 54 km/hr

Let's convert speed to m/sec:

$$\begin{aligned}54 \text { km/hr} = \Big(54 \times \frac{5}{18}\Big) \text{ m/sec} \\[1em] = \Big(3 \times \frac{5}{1}\Big) \text{ m/sec} \\[1em] = 15 \text{ m/sec}\end{aligned}$$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

Time = $\frac{300}{15}$ seconds = 20 seconds

Hence, the train will pass a bridge in 20 seconds.

Given:

Distance (D1) = Length of train = 760 m

Distance (D2) = Length of platform = 440 m

Total Distance = 760 m + 440 m = 1200 m

Time = 40 seconds

Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting the values in above, we get:

Speed = $\frac{1200}{40}$ m/sec = 30 m/sec

Let's convert speed to km/hr:

$$\begin{aligned}\therefore 30\text{ m/sec} = 30 \times \frac{18}{5} \text{ km/hr} \\[1em] = 6 \times \frac{18}{1} \text{ km/hr} \\[1em] = 6 \times 18 \text{ km/hr} \\[1em] = 108 \text{ km/hr}\end{aligned}$$

Hence, speed of the train is 108 km/hr.

Distance = Speed x Time is correct.

All other options are incorrect.

Hence, option 1 is the correct option.

To convert a speed of km/hr into m/sec, we multiply it by $\frac{5}{18}$.

Because, 1 km = 1000 m and 1 hr = 3600 sec.

$\frac{1000}{3600} = \frac{5}{18}$

Hence, option 3 is the correct option.

Given:

Speed = 90 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$:

$$\begin{aligned}90 \text { km/hr} = \Big(90 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(5 \times \frac{5}{1}\Big)\text{ m/sec} \\[1em] = (5 \times 5)\text{ m/sec} \\[1em] = 25 \text{ m/sec}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Speed = 30 km/hr, Distance = 200 m

Let's convert 30 km/hr to m/sec:

$$\begin{aligned}30 \text { km/hr} = \Big(30 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(5 \times \frac{5}{3}\Big)\text{ m/sec} \\[1em] = \Big(\frac{25}{3}\Big)\text{ m/sec}\end{aligned}$$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

$$\begin{aligned}\text {Time} = \frac{200}{\frac{25}{3}} \text {seconds} \\[1em] = 200 \times {\frac{3}{25}} \text{ seconds} \\[1em] = 8 \times {\frac{3}{1}} \text{ seconds} \\[1em] = 8 \times 3 \text{ seconds} \\[1em] = 24 \text{ seconds}\end{aligned}$$

Hence, option 2 is the correct option.

Given:

Speed = 48 km/hr, Time = 15 mins

Let's convert time to hrs:

1 hr = 60 mins

Time = 15 mins = $\frac{15}{60} \text{ hr} = \frac{1}{4} \text{ hr}$

Distance = Speed x Time

Substituting the values in above, we get:

Distance = $48 \times \frac{1}{4}$ km

Distance = 12 x 1 km = 12 km

Let's convert 12 km to meters:

1 km = 1000 m

Distance = 12 km = 12 x 1000 m = 12000 m

Hence, option 3 is the correct option.

Given:

Distance = 175 m, Time = 25 seconds

Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting the values in above, we get:

Speed = $\frac{175}{25}$ m/sec = 7 m/sec

Let's convert 7 m/sec to km/hr:

$$\begin{aligned}7 \text{ m/sec} = 7 \times \frac{18}{5} \text{ km/hr} \\[1em] = \frac{126}{5} \text{ km/hr} \\[1em] = 25.2 \text{ km/hr}\end{aligned}$$

Hence, option 4 is the correct option.

Given:

Distances: (D1) = 108 km, (D2) = 90 km

Time: (T1) = 2 hrs, (T2) = 1 hr

Total Distance = 108 km + 90 km = 198 km

Total Time = 2 hrs + 1 hr = 3 hrs

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{198}{3}$ km/hr = 66 km/hr

Hence, option 3 is the correct option.

(i) A speed of 81 km/hr expressed in metres/sec is 22.5 m/sec.

(ii) Which is greater — a speed of 50 m/sec or 50 km/hr? 50 m/sec.

(iii) If a moving body covers equal distances in equal intervals of time, its speed is said to be uniform.

(iv) In crossing a platform, a train has to cover a distance equal to the sum of the lengths of the train and the platform.

(v) To convert a speed of m/sec into km/hr, we multiply it by $\frac{18}{5}$.

Explanation

(i)

Given:

Speed = 81 km/hr

To convert km/hr to m/sec, multiply by $\frac{5}{18}$.

$$\begin{aligned}81 \text { km/hr}= \Big(81 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(9 \times \frac{5}{2}\Big)\text{ m/sec} \\[1em] = \frac{45}{2}\text{ m/sec} \\[1em] = 22.5 \text{ m/sec}\end{aligned}$$

Speed = 22.5 m/sec

(ii)

Let's convert 50 km/hr into m/sec:
$$\begin{aligned}50 \text { km/hr} = \Big(50 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \frac{250}{18}\text{ m/sec} \\[1em] ≈ 13.89 \text{ m/sec}\end{aligned}$$

Now compare 50 m/sec and 13.89 m/sec,

50 m/sec > 13.89 m/sec

So, 50 m/sec is greater.

(iii) The given statement is the definition of Uniform Speed (or constant speed). If the distance covered doesn't change for every second or minute that passes, the speed is consistent.

(iv) To completely clear a platform, the front of the train must enter the platform, travel the platform's length, and then the back of the train must travel the train's own length to fully exit.

(v) Since $1 \text{ m} = \frac{1}{1000}\text{ km}$ and $1 \text{ sec} = \frac{1}{3600}\text{ hr}$, the conversion factor simplifies to $\frac{3600}{1000} = \frac{18}{5}$.

(i) False
Reason —

To convert 45 km/hr to m/s, multiply by $\frac{5}{18}$.

$$\begin{aligned}45 \text { km/hr} = \Big(45 \times \frac{5}{18}\Big)\text{ m/s} \\[1em] = \Big(5 \times \frac{5}{2}\Big)\text{ m/s} \\[1em] = \frac{25}{2}\text{ m/s} \\[1em] = 12.5 \text{ m/s}\end{aligned}$$

Let's convert $12\frac{1}{3}$ m/s to decimal:

$12\frac{1}{3}\text{ m/s} = \frac{37}{3}\text{ m/s} = 12.33 \text{ m/s}$

12.5 m/s is not the same as $12\frac{1}{3} \text{ m/s } (12.33)\text{ m/s}$.

(ii) True
Reason —

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

This is the standard mathematical definition for Average Speed, used to calculate the speed over a journey with multiple parts.

(iii) False
Reason — If the distances are unequal, the speed is called Non-uniform or Variable speed. "Instantaneous speed" refers to the speed at a specific moment in time (like what a car's speedometer shows).

(iv) True
Reason — To convert km/hr to m/s, multiply by $\frac{5}{18}$.

Speed = 1 km/hr = $\Big(1 \times \frac{5}{18}\Big)\text{ m/s} = \frac{5}{18} \text{ m/s}$

(v) False
Reason —

Train length = 240 m

Platform length = 2 x 240 = 480 m

Total Distance = 240 m + 480 m = 720 m

Time = 54 sec

Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting values in the above, we get:

Speed = $\frac{720 \text{ m}}{54 \text{ sec}} = \frac{40}{3} \text{ m/s}$

Let's convert speed to km/hr:

$$\begin{aligned}\frac{40}{3} \text{ m/s} = \Big(\frac{40}{3} \times \frac{18}{5}\Big)\text{ km/hr} \\[1em] = (8 \times 6)\text{ km/hr} \\[1em] = 48 \text{ km/hr}\end{aligned}$$

(1)

Given:

Distance = 27 km = 27 x 1000 m = 27000 m

Time = 3 hours = 3 x 3600 sec = 10800 sec

Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting values in the above, we get:

Speed = $\frac{27000}{10800}\text{ m/sec} = \frac{270}{108}\text{ m/sec} = 2.5\text{ m/sec}$

Hence, option 2 is the correct option.

(2)

Given:

Distance = 27000 m

Time = 2 hours = 2 x 3600 sec = 7200 sec

Old Speed = 2.5 m/sec

New Speed = $\frac{\text{Distance}}{\text{Time}}$

Substituting values in the above, we get:

New Speed = $\frac{27000}{7200}\text{ m/sec} = \frac{270}{72}\text{ m/sec} = 3.75\text{ m/sec}$

Let's calculate the increase in speed:

Increase = New Speed - Old Speed

Increase = 3.75 m/sec - 2.5 m/sec = 1.25 m/sec

Hence, option 1 is the correct option.

(3)

Given:

Distance = Train length = 180 m

Speed = 54 km/hr

Let's convert speed to m/sec:

$$\begin{aligned}54 \text { km/hr} = \Big(54 \times \frac{5}{18}\Big)\text{ m/sec} \\[1em] = \Big(3 \times \frac{5}{1}\Big)\text{ m/sec} \\[1em] = 15 \text{ m/sec}\end{aligned}$$

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

Time = $\frac{180}{15}\text{ sec} = 12 \text{ sec}$

Hence, option 3 is the correct option.

(4)

Given:

Train length = 180 m

Platform length = 300 m

Total Distance = 180 m + 300 m = 480 m

Speed = 10 m/sec

Time = $\frac{\text{Total Distance}}{\text{Speed}}$

Substituting values in the above, we get:

Time = $\frac{480}{10}\text{ sec} = 48 \text{ sec}$

Time taken by the train to pass the platform is 48 sec.

Therefore, none of the provided options is correct for the given values.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

Given:

Speed = 48 km/hr

Distance = 28 km

Time = 35 minutes

Let's convert time to hours:

1 hour = 60 minutes

Time = 35 minutes = $\frac{35}{60}\text{ hr} = \frac{7}{12}\text{ hr}$

Distance = Speed x Time

Substituting the values in above, we get:

Distance = $\Big(48 \times \frac{7}{12}\Big)\text{ km} = \Big(4 \times \frac{7}{1}\Big)\text{ km} = 28 \text{ km}$

The Assertion is True.

Reason:

Distance = speed x time

This is the correct, standard formula for distance.

Hence, option 1 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

Given:

Distance = 12 km

Onward speed = 6 km/hr

Return speed = 4 km/hr

Let's find total time:

Onward time (T1) = $\frac{\text{Distance}}{\text{Onward speed}} = \frac{12}{6}\text{ hrs} = 2 \text{ hrs}$

Return time (T2) = $\frac{\text{Distance}}{\text{Return speed}} = \frac{12}{4}\text{ hrs} = 3 \text{ hrs}$

Total Time = 2 hrs + 3 hrs = 5 hrs

Total Distance = 12 km (to city) + 12 km (back home) = 24 km

Average speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average speed = $\frac{24}{5}\text{ km/hr} = 4.8 \text{ km/hr}$.

The Assertion is True.

Reason:

Average speed = $\frac{\text{Onward speed + return speed}}{2}$

This is False. You cannot simply find the average of the speeds. You must always use $\frac{\text{Total Distance}}{\text{Total Time}}$.

The reason is false.

Hence, option 3 is the correct option.

Let the actual distance travelled by the person be d km.

Let the actual time taken be t hours.

Case 1 (Actual): Speed = 10 km/hr, Distance = d km

Time = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

t = $\frac{d}{10}$ hours ........(i)

Case 2 (New): Speed = 14 km/hr, Distance = (d + 20) km

In the same time t,

Distance = Speed × Time

Substituting the values, we get:

d + 20 = 14 × t

d + 20 = 14 × $\frac{d}{10} \quad$ [From (i)]

d + 20 = $\frac{14d}{10}$

d + 20 = $\frac{7d}{5}$

5(d + 20) = 7d

5d + 100 = 7d

⇒ 7d - 5d = 100

⇒ 2d = 100

⇒ d = 50 km

Hence, option 1 is the correct option.

Given:

Speed of sound = 300 m/s

Time taken by person 1 to hear the sound (T₁) = 5 seconds

Time taken by person 2 to hear the sound (T₂) = 6 seconds

Distance = Speed x Time

Let's find the distance of each person from the source of the sound:

Distance of person 1 from source (D₁) = 300 x 5 = 1500 m = 1.5 km

Distance of person 2 from source (D₂) = 300 x 6 = 1800 m = 1.8 km

Maximum distance occurs when the two persons are on opposite sides of the source.

Maximum distance = D₁ + D₂ = 1.5 km + 1.8 km = 3.3 km

Minimum distance occurs when the two persons are on the same side of the source.

Minimum distance = D₂ - D₁ = 1.8 km - 1.5 km = 0.3 km

Hence, option 4 is the correct option.

Given:

Initial distance between thief and policeman = 200 m

Speed of thief = 10 km/hr

Speed of policeman = 11 km/hr

Time = 6 minutes

Since both are running in the same direction, the policeman gains on the thief at the relative speed.

Relative speed = Speed of policeman - Speed of thief

Relative speed = 11 km/hr - 10 km/hr = 1 km/hr

Let's convert relative speed to m/min:

1 km/hr = $\frac{1000 \text{ m}}{60 \text{ min}} = \frac{50}{3}$ m/min

Distance gained by policeman in 6 minutes

Distance = Speed x Time

Substituting the values, we get:

Distance gained = $\Big(\frac{50}{3} \times 6\Big)\text{ m} = \frac{300}{3}\text{ m} = 100 \text{ m}$

Remaining distance between them = Initial distance - Distance gained

Remaining distance = 200 m - 100 m = 100 m

Hence, option 1 is the correct option.

Given:

Distance: (D₁) = 108 km, (D₂) = 90 km

Time: (T₁) = 2 hrs, (T₂) = 1 hr

Total Distance = 108 km + 90 km = 198 km

Total Time = 2 hrs + 1 hr = 3 hrs

Average Speed = $\frac{\text{Total Distance}}{\text{Total Time}}$

Substituting the values in above, we get:

Average Speed = $\frac{198}{3}$ km/hr = 66 km/hr

Hence, option 3 is the correct option.

Given:

Total Distance = 6 km

Total Time = 84 minutes

Let's convert total time to hours:

1 hour = 60 minutes

∴ 84 minutes = $\frac{84}{60}\text{ hr} = \frac{7}{5}\text{ hr}$

Distance covered in first part = $\frac{2}{3} \times 6$ km = 4 km

Speed for first part = 4 km/hr

Time taken for first part (T₁) = $\frac{\text{Distance}}{\text{Speed}}$

Substituting the values in above, we get:

T₁ = $\frac{4}{4}$ hr = 1 hr

Remaining distance = 6 km - 4 km = 2 km

Remaining time (T₂) = Total Time - T₁

T₂ = $\frac{7}{5}$ hr - 1 hr = $\frac{7 - 5}{5}$ hr = $\frac{2}{5}$ hr

Speed for remaining distance = $\frac{\text{Remaining Distance}}{\text{Remaining Time}}$

Substituting the values in above, we get:

$$\begin{aligned}\text{Speed} = \frac{2}{\frac{2}{5}}\text{ km/hr} \\[1em] = \Big(2 \times \frac{5}{2}\Big)\text{ km/hr} \\[1em] = 5 \text{ km/hr}\end{aligned}$$

Hence, option 1 is the correct option.