Question

A particle moves in a straight line, so that $s=\sqrt{t}$ then its acceleration is proportional to

• A. $velocity$
• B. ${\left(velocity\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• C. ${\left(velocity\right)}^3$
• D. ${\left(velocity\right)}^2$

Answer 1

The correct option is C

${\left(velocity\right)}^3$

Step 1: Given data.

Distance covered, $s=\sqrt{t}$

Step 2: Finding the acceleration of the particle.

We know that,

$a=\frac{dv}{dt}$ ….$\left(i\right)$

Where $a$ is acceleration and $\frac{dv}{dt}$ is change in velocity with respect to time.

$v=\frac{ds}{dt}$

Where $v$ is acceleration and $\frac{ds}{dt}$ is change in distance with respect to time.

Now,

$v=\frac{ds}{dt}$

$v=\frac{d\sqrt{t}}{dt}$

$v=\frac{1}{2}{t}^{-\frac{1}{2}}$

Differentiate equation $\left(ii\right)$ with respect to time we get.

$\frac{dv}{dt}=-\frac{1}{4}{t}^{-\frac{3}{2}}$

$\frac{d^{2}s}{d{t}^2}=-\frac{1}{4}{t}^{-\frac{3}{2}}$ $\left[\because v=\frac{ds}{dt}\right]$

Thus,

$a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}=-\frac{1}{4}{t}^{-\frac{3}{2}}$

$a=\frac{dv}{dt}=\frac{d^{2}s}{d{t}^2}=-\frac{1}{4}{t}^{-\frac{3}{2}}$

$a=-\frac{1}{4}{\left(t^{-\frac{1}{2}}\right)}^3$

$a=-\frac{1}{4}{\left(2v\right)}^3$ $\left[\because v=\frac{1}{2}{t}^{-\frac{1}{2}}\Rightarrow 2v=t^{-\frac{1}{2}}\right]$

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

Therefore,

$a\propto v^3$

Hence, option $\mathit{C}$ is correct.