The relation between time $t$ and distance $x$ is $t=a x^{2}+b x$ where $a$ and $b$ are constants. The acceleration is

JEE Main · 2005 · 08 May 2005 · Offline · Kinematics

The relation between time

t

and distance

x

t = a x^{2} + b x

where

a

and

b

are constants.

The acceleration is

(A) $- 2 ab v^{2}$
(B) $2 b v^{3}$
(C) $- 2 a v^{3}$ ✓ Correct
(D) $2 a v^{2}$

Answer: (C)

Solution

Differentiate

t = a x^{2} + b x

with respect to time:

1 = (2 a x + b) \frac{d x}{d t} = (2 a x + b) v

, so

2 a x + b = \frac{1}{v}

. Differentiating once more,

0 = 2 a v^{2} + (2 a x + b) \frac{d v}{d t}

. Replacing

2 a x + b

\frac{1}{v}

gives

0 = 2 a v^{2} + \frac{1}{v} \frac{d v}{d t}

, so the acceleration is

\frac{d v}{d t} = - 2 a v^{3}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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