Kinematics

JEE Main level, numerical answer, medium

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Question 2 of 308

A vehicle travelling at 20 m/s20\,\text{m/s} stops in a distance of 40 m40\,\text{m} upon applying brakes. If the speed is doubled to 40 m/s40\,\text{m/s}, the stopping distance under same retardation is (in m):

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Answer: 160

Stopping distance s=u22a∝u2s = \frac{u^2}{2a} \propto u^2. When speed is doubled, distance increases by 22=42^2 = 4 times: s′=4×40=160 ms' = 4 \times 40 = 160\,\text{m}.

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