JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Kinematics

A car is moving with speed of

150 km / h

and after applying the brake it will move 27 m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ______ m distance.

Answer: 3

Solution

The stopping distance is given by

d = \frac{v ^{2}}{2 a}

, where

v

is the initial speed and

a

is the braking acceleration (constant). When the speed is reduced to

v /3

, the new stopping distance is

d^{'} = \frac{( v /3 ) ^{2}}{2 a} = \frac{v ^{2}}{9 \times 2 a} = \frac{d}{9}

. Given

d = 27

m, we have

d^{'} = \frac{27}{9} = 3

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer

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