Speed, Distance & Time — Exercise 12(B)

A train covers 200 km in the first two hours, then 126 km in the next 2 hours and finally 143 km in the last 3 hours. Find the average speed of the train for the whole of this journey.

Answer

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

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

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

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

A bus travels at a speed of 72 km/h for 5 hours and at a speed of 90 km/h for 4 hours. Find the average speed of the bus for the whole journey.

Answer

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 = $\dfrac{720 \text{ km}}{9 \text{ h}} = 80 \text{ km/h }$

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

Out of a distance of 80 km, the first 60 km is covered at a speed of 40 km/h and the remaining distance at a speed of 20 km/h. Calculate the average speed.

Answer

For the first part :

Distance = 60 km and speed = 40 km/h

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

For the remaining part :

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

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

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

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

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

P and Q are two stations. A car goes from station P to station Q at a speed of 60 km/h and returns back at a speed of 30 km/h. Find the average speed for the entire journey.

Answer

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

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

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

Total distance covered = x + x = 2x km

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

Calculating average speed,

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

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

Out of a journey of 300 km; the first part of distance 85 km is covered at a speed of 51 km/h, the second part of 90 km at a speed of 135 km/h and the remaining distance at a speed of 75 km/h. Find:

(i) the distance covered at a speed of 75 km/h.

(ii) the total time taken.

(iii) the average speed for the whole journey.

Answer

(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 = $\dfrac{85 \text{ km}}{51 \text{ km/h}} = \dfrac{5}{3} \text{ h }$

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

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

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

Hence, the total time taken = 4 hours.

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

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

A motorcycle covers a distance of 72 km at a speed of 36 km/h and a distance of 135 km at a speed of 45 km/hr. Find the average speed of the motorcycle.

Answer

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

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

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

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

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

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

Speed of car P is 120 km/h and speed of car Q is 75 km/h.

(i) If both are moving in opposite directions, what is their relative speed?

(ii) What is their relative speed when they are moving in the same direction?

Answer

(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.

A train 900 m long, crosses a pole in 45 sec. Find its speed in km per hour.

Answer

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

Distance = 900 m and time = 45 s

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

Converting this speed into km/h :

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

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

Find the length of the train moving at a speed of 90 km/h, if it passes a standing man in 8 seconds.

Answer

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

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

Time = 8 s

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

Hence, the length of the train = 200 m.

A 100 m long train passes a 200 m long platform in 20 seconds. Find the speed of the train.

Answer

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

Converting this speed into km/h :

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

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

Two cars start from the same place with speeds 80 km/h and 50 km/h. Find the distance between the two cars at the end of 3 hours, if:

(i) they are going in the same direction.

(ii) they are going in the opposite directions.

Answer

(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.

A train, 80 m long, passes a platform 220 m long. If the speed of the train is 45 km/h, find the time taken by the train.

Answer

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

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

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

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

The speed of a bus is 90 km/h and the speed of a truck is 72 km/h. Both start from the same place. Find the distance between the two after 20 seconds, when they go in the:

(i) same direction

(ii) opposite directions.

Answer

Converting both the speeds into m/s :

1 km = 1000 m and 1 h = 3600 s

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

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

(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.

A train passes a 50 m long railway platform in $4\dfrac{1}{2}$ seconds and a pole in 2 seconds. Find the length of the train and its speed.

Answer

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.

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

Subtracting equation (1) from equation (2) :

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

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

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

Converting the speed into km/h :

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

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

A train passes a platform, 225 m length, in 21 sec and a man, standing on the platform, in 6 sec. Find:

(i) the length of the train.

(ii) the speed of the train.

Answer

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) :

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

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 :

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

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