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MHT CET 2020 (16 Oct, Shift 1) · Equations of Kinematics

Equations of Kinematics question from MHT CET 2020 (16 Oct, Shift 1)

Q29MHT CET 2020 · 16 OCT · SHIFT 1Easy

Equations of Kinematics

A moving body is covering distances which are proportional to square of the time. Then, the acceleration of the body is

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Answer: (c)

$s \propto t^2$ means $s = kt^2$ .
$v = \frac{ds}{dt} = 2kt$ , $a = \frac{dv}{dt} = 2k$
So acceleration is constant and non-zero. So the answer is (c).

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