question_answer

The relation between time t and distance\[x\]is \[t=a{{x}^{2}}+bx,\]where a and b are constants. The acceleration is                         [AIEEE 2005]

A) \[-2ab{{v}^{2}}\]

B) \[2b{{v}^{3}}\]

C) \[-2a{{v}^{3}}\]

D) \[2a{{v}^{2}}\]

Correct Answer: C

Solution :

[c] Given, \[t=a{{x}^{2}}+bx\]

Differentiating w.r.t. t

\[\frac{d}{dt}=2ax\frac{dx}{dt}+b\frac{dx}{dt}\]

\[v=\frac{dx}{dt}=\frac{1}{(2ax+b)}\]

Again differentiating w.r.t. t

\[\frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{-2a}{{{(2ax+b)}^{2}}}\frac{dx}{dt}\]

\[\therefore f=\frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{-2a}{{{(2ax+b)}^{2}}}.\frac{1}{(2ax+b)}\]

or         \[f=\frac{-2a}{{{(2ax+b)}^{3}}}\]

\[\therefore f=-2a{{v}^{3}}\]