Q. The relation between time $t$ and distance $x$ is $t = ax^{2} + bx,$ where $a\ and\ b$constants are. The acceleration is
A) $- 2abv^{2}$
B) $2\ bv^{3}$
C) $- 2av^{3}$
D) $2av^{2}$
Explanation
(c) Given, $t = ax^{2} + bx$
Differentiating w.r.t. $t$
$\frac{dx}{dt} = 2\ ax\frac{dx}{dt} + b\frac{dx}{dt}$
$v = \frac{dx}{dt} = \frac{1}{(2ax\ + b)}$
Again differentiating w.r.t. $t$
$\frac{d^{2}x}{dt^{2}} = \frac{- 2a}{(2\ ax + b)^{2}}\frac{dx}{dt}$
$\therefore f = \frac{d^{2}x}{dt^{2}} = \frac{- 1}{(2\ ax + b)^{2}}\ .\frac{2a}{(2ax + b)}$
Or $f = \frac{- 2a}{(2ax\ + \ b)^{3}}$
$\therefore f = - 2av^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.