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Chapter 2 Motion In One Dimension | Exercise 2C

Question 18

A car travels with a uniform velocity of 25m/s for 5s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10s. Find: (i) the distance which the car travels before the brakes are applied, (ii) the retardation and (iii) the distance travelled by car after applying the brakes.

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Given: u=25m/s; v=0;

(i) The distance which the car travels before brakes are applied is:

Distance = speed x time

= 25 x 5

= 125m

(ii) Retardation = (final velocity-initial velocity)/time taken

= -(5/2) = -2.5m/s2

(iii) To find distance travelled by the car after applying brakes

Time taken by the car to stop after applying brakes =10s

Assume ‘s’ to be the distance the car travels after brakes are applied

We know that v^2 - u^2 = 2as

(0 x 0) - (25 x 25)

= 2(-2.5)(s)

– 625 = -5s

s=125m

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Question 18

A car travels with a uniform velocity of 25m/s for 5s. The brakes are then applied and the car is uniformly retarded and comes to rest in further 10s. Find: (i) the distance which the car travels before the brakes are applied, (ii) the retardation and (iii) the distance travelled by car after applying the brakes.

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given: u=25m/s; v=0;

(i) The distance which the car travels before brakes are applied is:

Distance = speed x time

= 25 x 5

= 125m

(ii) Retardation = (final velocity-initial velocity)/time taken

= -(5/2) = -2.5m/s2

(iii) To find distance travelled by the car after applying brakes

Time taken by the car to stop after applying brakes =10s

Assume ‘s’ to be the distance the car travels after brakes are applied

We know that v^2 - u^2 = 2as

(0 x 0) - (25 x 25)

= 2(-2.5)(s)

– 625 = -5s

s=125m

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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