Question

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

Answer 1

S = u^2 / (2a)

Same car will cause same retardation (a) in both cases.

∴ S ∝ u^2

S1 / S2 = (u1 / u2)^2
S2 = S1 * (u2 / u1)^2
S2 = 8 * (60 / 30)^2
S2 = 32 m

It will come to rest after 32 m

Answer 2

Answer:

32m

Explanation:

v² = u² + 2as…….. where a: retardation (opposite of acceleration) and s: distance covered.

1st case:

u = 30 km/h = 25/3 m/s

v = 0,

s = 8 m

Therefore, 0² = (25/3)² - 2 × a × 8

This gives a = 4.34 m/s²

2nd case:

u = 60 km/h = 50/3 m/s

v = 0

a = 4.34 m/s²

Now, 0² = (50/3)² - 2 × 4.34 × s

This gives s = 32 m

Thus on applying brakes the car will stop after covering 32 m.