Question

The velocity of a particle moving in the positive direction of the x-axis varies as $V=\alpha \sqrt{x}$ where $\alpha$ is a positive constant. Assuming that at the moment $t=0$ the particle was locate at the point $x=0$, find velocity at $t=2s$.

• A. ${\alpha }^2$
• B. ${\alpha }^2/2$
• C. $\alpha$
• D. None of these

Answer 1

Answer: (A)

${\alpha }^2$

Velocity of the particle $\frac{dx}{dt}=\alpha \sqrt{x}$

Rearranging the terms we get $\frac{dx}{\sqrt{x}}=\alpha dt$

Integrating both sides, $\int \frac{dx}{\sqrt{x}}=\alpha \int dt$

We get $2\sqrt{x}=\alpha t$.

$\therefore$ $x=\frac{{\alpha }^2{t}^2}{4}$

Differentiating w.r.t time we get velocity of the particle as a function of time,

$v=\frac{dx}{dt}=\frac{{\alpha }^{2}t}{2}$

At time $t=2$ s,

$v=\frac{{\alpha }^2\times 2}{2}={\alpha }^2$