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
${\int }_0^x\frac{dx}{\sqrt{x}}={\int }_0^1\alpha dt$
$\left[\because att=0,x=0andletatanytimet,particleisatx\right]$
$\Rightarrow \frac{x^{1/2}}{1/2}{|}_0^x=\alpha t$
$\Rightarrow x^{1/2}=\frac{\alpha }{2}t$
$\Rightarrow x=\frac{{\alpha }^2}{4}\times t^2\Rightarrow x\propto t^2$