Question

A particle located at $x=0$ at time $t=0$, starts moving along the positive $x-$direction with a velocity $v$ that varies as $v=\alpha \sqrt{x}.$ The displacement of the particle varies with time as

• A. $t^2$
• B. $t$
• C. $t^{1/2}$
• D. $t^3$

Answer 1

The correct option is A $t^2$

Given, $v=\alpha \sqrt{x}$
As we know that, $v=\frac{dx}{dt}$, where, $x$ is the displacement of the particle.
$\Rightarrow \frac{dx}{dt}=\alpha \sqrt{x}$
$\Rightarrow {\int }_0^x\frac{dx}{\sqrt{x}}={\int }_0^t\alpha dt$
$\Rightarrow {\left[\frac{x^{1/2}}{1/2}\right]}_0^x=\alpha t$
$\Rightarrow \sqrt{x}=\frac{\alpha t}{2}$
$\Rightarrow x=\frac{{\alpha }^2{t}^2}{4}$
$\Rightarrow x\propto t^2$
Hence, displacement will varies with time as $x\propto t^2$