Question

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

• A. $-2ab{v}^2$
• B. $2b{v}^3$
• C. $-2a{v}^3$
• D. $2a{v}^2$

Answer 1

Answer: (C)

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