Question

The relation between time $t$ and distance $x$ is $t = ax^2 + bx$, where $a$ and $b$ are constants. The acceleration is:

• A. $-2abv^2$
• B. $2bv^3$
• C. $-2av^3$
• D. $2av^2$

Answer 1

Answer: (C)

Given $t = ax^2 + bx$
Differentiating $w.r.t. \,t$
$\frac{dt}{dt}=2\,ax \frac{dt}{dt}+b \frac{dx}{dt}$
$v=\frac{dx}{dt}=\frac{1}{\left(2\,ax+b\right)}$
Again differentiating, $w.r.t. \,t$
$\frac{d^{2}x}{dt^{2}}=\frac{-2a}{\left(2\,ax+b\right)^{2}}.\frac{dx}{dt}$
$\therefore f=\frac{d^{2}x}{dt^{2}}$
$=-\frac{-1}{\left(2\,ax+b\right)^{2}}. \frac{2a}{\left(2\,ax+b\right)}$
or $f=\frac{-2a}{\left(2\,ax+b\right)^{3}}$
$\therefore f=-2\,av^{3}$