JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Kinematics

The relation between time

t

and distance

x

for a moving body is given as

t = m x^{2} + n x

, where

m

and

n

are constants. The retardation of the motion is : (Where

v

stands for velocity)

(A) $2 n^{2} v^{2}$
(B) $2 mn v^{3}$
(C) $2 m v^{3}$ ✓ Correct
(D) $2 n v^{3}$

Answer: (C)

Solution

From

t = m x^{2} + n x

, differentiate with respect to

x

\frac{d t}{d x} = 2 m x + n

. Since

v = \frac{d x}{d t}

, we have

v = \frac{1}{2 m x + n}

. The acceleration is

a = v \frac{d v}{d x} = v \cdot \frac{- 2 m}{( 2 m x + n ) ^{2}} = - 2 m v^{3}

. The retardation (magnitude) is

2 m v^{3}

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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