A car travelling at a speed of30km/h is bought to rest in a distance of 8m by applying brakes. If the same car is moving at a speed of 60km/h the in it can be brought to rest with same brakes in???

Answer 1

everything after "60 km/h" is just gibberish.

Answer 2

Assuming that you mean:

"A car travelling at a speed of 30km/h is bought to rest in a distance of 8m by applying brakes. If the same car is moving at a speed of 60km/h, then the distance in which it can be brought to rest with the same brakes is?"

In the initial case,
v = 0
u = 30km/hr = 8.33m/s
s = 8m

Using 2as = v^2 - u^2,
a = - (8.33^2/2*8)
= -69.444/16
= -4.34

Hence, this car's brakes always provide a deceleration of -4.34m/s^2

Coming to the second case,
a = -4.34m/s^2
v = 0
u = 60km/hr = 16.66m/s

Using the same identity,

s = -u^2/2*a
= -277.666/8.68
= 31.91m
= 32m

Answer 3

You could also directly say using the same equation that the stopping distance varies with the square of the initial speed, and hence when the initial speed is doubled, the stopping distance is four times as much.

Answer 4

Arora's last statement about the relation of distance related to the square of the speed is the point of this conceptual question. Variants of this question often appear in college advanced placement exams. Double speed, distance to stop doubles^2 (quadruples).