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MHT CET 2025 (26 Apr, Shift 2) · Equations of Kinematics

Equations of Kinematics question from MHT CET 2025 (26 Apr, Shift 2)

Q4MHT CET 2025 · 26 APR · SHIFT 2Medium

Equations of Kinematics

A car moving at a speed ' v ' is stopped in a certain distance when the brakes produce a deceleration ' $a$ '. If the speed of the car is ' $nv$ ', what must be the deceleration of the car to stop it in the same distance and in the same time?

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Answer: (c)

Stopping distance $s = \frac{v^2}{2a}$ .
Same distance with speed $nv$ : $\frac{(nv)^2}{2a'} = \frac{v^2}{2a} \Rightarrow a' = n^2a$ .
So the answer is (c) $n^2 \cdot a$ .

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