Question

The relation between time $t$ and distance $x$ for a moving body is given as $t=m{x}^2+nx$, where $m$ and $n$ are constants. The retardation of the motion is :
$\left(\text{Where}v\text{stands for velocity}\right)$

• A. $2m{v}^3$
• B. $2mn{v}^3$
• C. $2n{v}^3$
• D. $2n^2{v}^3$

Answer 1

The correct option is A $2m{v}^3$

Given that,

$t=m{x}^2+nx$

Differentiating w.r.t. $x$, we get

$\frac{dt}{dx}=\frac{1}{v}=2mx+n$

$v=\frac{1}{2mx+n}.......\left(1\right)$

Again differentiating w.r.t. $t$, we get

$a=\frac{dv}{dt}=-\frac{2m}{(2mx+n{)}^2}\left(\frac{dx}{dt}\right)$

Substituting $\left(1\right)$ in above equation,

$a=-\left(2m\right)v^3$

So, the retardation will be,

$-a=2m{v}^3$

Hence, option $\left(\text{A}\right)$ is correct.