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

Answer: (A)

$v=\alpha\sqrt{x}$
$\frac{dx}{dt}=\alpha\sqrt{x}\,\left(\because v=\frac{dx}{dt}\right)$
$\frac{dx}{\sqrt{x}}=\alpha\,dt$
Perform integration
$\int^{x}_{0} \frac{dx}{\sqrt{x}}=\int^{i}_{0}\,\alpha\,dt$
[$\because$ at $t = 0, x = 0$ and let at any time $t$, particle is at $x$]
$\Rightarrow \frac{x^{-1/2}}{-1/2}|^x_0=\alpha t$
$\Rightarrow x^{-1/2}=\frac{\alpha}{2}t$
$\Rightarrow x=\frac{\alpha^{2}}{4}\times t^{2} \Rightarrow x\,\propto\,t^{2}$