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JEE Main 2018 (15 Apr, Shift 1) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2018 (15 Apr, Shift 1)

Q32JEE Main 2018 · 15 APR · SHIFT 1Easy

Equations of Kinematics

An automobile, travelling at $40$ km/h, can be stopped at a distance of $40$ m by applying brakes. If the same automobile is travelling at $80$ km/h, the minimum stopping distance, in metres, is (assume no skidding):

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Answer: (d)

For the same retardation, $s = \frac{u^2}{2a} \Rightarrow s \propto u^2$ .
Speed doubles, so $s' = 4 \times 40 = 160\text{ m}$ .
So the answer is (d) $160$ m.

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