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JEE Main 2021 (25 Jul, Shift 2) · Velocity, Acceleration & Variable Motion

Velocity, Acceleration & Variable Motion question from JEE Main 2021 (25 Jul, Shift 2)

Q26JEE Main 2021 · 25 JUL · SHIFT 2Medium

Velocity, Acceleration & Variable Motion

The relation between time t and distance x for a moving body is given as t = mx2 + nx, where m and n are constants. The retardation of the motion is: (When v stands for velocity)

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Answer: (a)

$\dfrac{dt}{dx} = 2mx + n = \dfrac{1}{v}$
Differentiate w.r.t. $t$ : $2mv = -\dfrac{1}{v^2}\dfrac{dv}{dt} \Rightarrow a = -2mv^3$
Retardation $= 2mv^3$
So the answer is (a).

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