Idea of Speed, Distance & Time — Exercise 11(B)

A train 180 m long is running at a speed of 90 km/h. How long will it take to pass a railway signal?

Answer

Given,

Length of the train = 180 m and speed = 90 km/h.

To pass a railway signal (a stationary object), the distance to be covered = length of the train = 180 m.

Converting the speed into m/s (multiply by $\dfrac{5}{18}$) :

Speed = 90 × $\dfrac{5}{18} = \dfrac{450}{18} = 25$ m/s.

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{180}{25} = 7.2$ seconds.

Hence, the train takes 7.2 seconds to pass the railway signal.

A train whose length is 150 m, passes a telegraph pole in 10 s. Find the speed of the train in km/h.

Answer

Given,

Length of the train = 150 m and time taken to pass the telegraph pole = 10 s.

To pass a telegraph pole (a stationary object), the distance covered = length of the train = 150 m.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{150}{10} = 15$ m/s.

Converting the speed into km/h (multiply by $\dfrac{18}{5}$) :

Speed = 15 × $\dfrac{18}{5} = \dfrac{270}{5} = 54$ km/h.

Hence, the speed of the train = 54 km/h.

A train 120 m long passes a railway platform 160 m long in 14 s. How long will it take to pass another platform which is 100 m long?

Answer

Given,

Length of the train = 120 m.

When the train passes the first platform (160 m long) :

Distance covered = length of the train + length of the platform = 120 m + 160 m = 280 m in 14 s.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{280}{14} = 20$ m/s.

When the train passes the second platform (100 m long) :

Distance to be covered = length of the train + length of the platform = 120 m + 100 m = 220 m.

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{220}{20} = 11$ seconds.

Hence, the train takes 11 seconds to pass the platform which is 100 m long.

Mr. Amit can walk 8 km in 1 hour 20 minutes.

(a) How far does he go in:

(i) 10 minutes?

(ii) 30 seconds?

(b) How long will it take him to walk:

(i) 2500 m?

(ii) 6.5 km?

Answer

Given,

Mr. Amit walks 8 km in 1 hour 20 minutes.

Distance = 8 km = 8000 m.

Time = 1 hour 20 minutes = 80 minutes.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{8000}{80} = 100$ m/min.

(a) (i) Distance covered in 10 minutes = Speed × Time = 100 × 10 = 1000 m = 1 km.

(ii) Speed in m/s = $\dfrac{100}{60} = \dfrac{5}{3}$ m/s.

Distance covered in 30 seconds = Speed × Time = $\dfrac{5}{3}$ × 30 = 50 m.

Hence, he goes 1 km in 10 minutes and 50 m in 30 seconds.

(b) (i) Time taken to walk 2500 m = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{2500}{100} = 25$ minutes.

(ii) 6.5 km = 6.5 × 1000 m = 6500 m.

Time taken to walk 6.5 km = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{6500}{100} = 65$ minutes = 1 hour 5 minutes.

Hence, he takes 25 minutes to walk 2500 m and 1 hour 5 minutes (65 minutes) to walk 6.5 km.

Which is greater: a speed of 45 km/h or a speed of 12.25 m/sec? How much is the distance travelled by each in 2 seconds?

Answer

To compare the two speeds, convert both into the same unit (m/s).

Converting 45 km/h into m/s (multiply by $\dfrac{5}{18}$) :

45 km/h = 45 × $\dfrac{5}{18} = \dfrac{225}{18} = 12.5$ m/s.

Since 12.5 m/s > 12.25 m/s, a speed of 45 km/h is greater.

Distance travelled in 2 seconds :

At 12.5 m/s : Distance = Speed × Time = 12.5 × 2 = 25 m.

At 12.25 m/s : Distance = Speed × Time = 12.25 × 2 = 24.5 m.

Hence, a speed of 45 km/h is greater. In 2 seconds, an object moving at 45 km/h covers 25 m, whereas an object moving at 12.25 m/s covers 24.5 m.

A and B start from the same point and at the same time with speeds 15 km/h and 12 km/h respectively. Find the distance between A and B after 6 hours if both move in:

(i) same direction

(ii) the opposite directions.

Answer

Given,

Speed of A = 15 km/h, speed of B = 12 km/h and time = 6 hours.

Distance covered by A in 6 hours = Speed × Time = 15 × 6 = 90 km.

Distance covered by B in 6 hours = Speed × Time = 12 × 6 = 72 km.

(i) When both move in the same direction, the distance between them = difference between the distances covered.

Distance between A and B = 90 km − 72 km = 18 km.

Hence, in the same direction the distance between A and B = 18 km.

(ii) When both move in opposite directions, the distance between them = sum of the distances covered.

Distance between A and B = 90 km + 72 km = 162 km.

Hence, in opposite directions the distance between A and B = 162 km.

A and B start from the same place, in the same direction and at the same time with speeds 6 km/h and 2 m/sec respectively. After 5 hours who will be ahead and by how much?

Answer

Given,

Speed of A = 6 km/h, speed of B = 2 m/sec and time = 5 hours.

Converting B's speed into km/h (multiply by $\dfrac{18}{5}$) :

Speed of B = 2 × $\dfrac{18}{5} = \dfrac{36}{5} = 7.2$ km/h.

Distance covered by A in 5 hours = Speed × Time = 6 × 5 = 30 km.

Distance covered by B in 5 hours = Speed × Time = 7.2 × 5 = 36 km.

As both move in the same direction, the distance between them = difference between the distances covered = 36 km − 30 km = 6 km.

Since B covers the greater distance, B is ahead.

Hence, B will be ahead of A by 6 km.

Mohit covers a certain distance in 6 hours by his scooter at a speed of 40 km h^-1.

(i) Find the time taken by Manjoor to cover the same distance by his car at the speed of 60 km h^-1.

(ii) Find the speed of Joseph, if he takes 8 hours to complete the same distance.

Answer

Given,

Mohit covers the distance in 6 hours at a speed of 40 km/h.

Distance = Speed × Time = 40 × 6 = 240 km.

(i) Time taken by Manjoor at 60 km/h = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{240}{60} = 4$ hours.

Hence, Manjoor takes 4 hours to cover the same distance.

(ii) Speed of Joseph (who takes 8 hours) = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{240}{8} = 30$ km/h.

Hence, the speed of Joseph = 30 km/h.

A boy swims 200 m in still water and then returns back to the point of start in total 10 minutes. Find the speed of his swim in (i) m s^-1 (ii) km h^-1.

Answer

Given,

The boy swims 200 m and then returns, so total distance = 200 m + 200 m = 400 m.

Total time taken = 10 minutes

= 10 × 60 seconds

= 600 seconds.

(i) Speed = $\dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{400}{600} = \dfrac{2}{3}$ m/s.

Hence, his swimming speed = $\dfrac{2}{3}$ m/s.

(ii) Converting the speed into km/h (multiply by $\dfrac{18}{5}$) :

Speed = $\dfrac{2}{3}$ × $\dfrac{18}{5} = \dfrac{36}{15} = \dfrac{12}{5} = 2.4$ km/h.

Hence, his swimming speed = 2.4 km/h.

A distance of 14.4 km is covered in 2 hours 40 minutes. Find the speed in m s^-1. With this speed Sakshi goes to her school, 240 m away from her house and then returns back. How much time, in all, will Sakshi take?

Answer

Given,

Distance = 14.4 km and time = 2 hours 40 minutes.

Converting into metres and seconds :

Distance = 14.4 km

= 14.4 × 1000 m

= 14400 m.

Time = 2 hours 40 minutes

= (2 × 3600 + 40 × 60) seconds

= (7200 + 2400) seconds

= 9600 seconds.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{14400}{9600} = 1.5$ m/s.

Sakshi goes 240 m to school and returns, so total distance = 240 m + 240 m = 480 m.

Total time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{480}{1.5} = 320$ seconds.

Total time taken = 5 minutes 20 seconds.

Hence, the speed = 1.5 m/s and Sakshi takes 320 seconds (5 minutes 20 seconds) in all.