Question

Consider a particle moving along the positive direction of $X$ -axis. The velocity of the particle is given by $v=\alpha \sqrt{x}(\alpha$ is a positive constant). At time $t=0$, if the particle is located at $x=0$, the time dependence of the velocity and the acceleration of the particle are respectively

• A. $\frac{{\alpha }^2}{2}t$ and $\frac{{\alpha }^2}{2}$
• B. ${\alpha }^{2}t$ and ${\alpha }^2$
• C. $\frac{\alpha }{2}t$ and $\frac{\alpha }{2}$
• D. $\frac{{\alpha }^2}{4}t$ and $\frac{{\alpha }^2}{4}$

Answer 1

Answer: (A)

Given, $v=\alpha \sqrt{x}$ and at $x=0,t=0$
To find velocity and acceleration in terms of time, we need to find $x$ as function of $t$.
$\because \frac{dx}{dt}=\alpha \sqrt{x}$
$\therefore \frac{dx}{\sqrt{x}}=\alpha dt$
On integrating both sides, we get
$2\sqrt{x}=\alpha t+C$
Using, $x=0$ at $t=0\Rightarrow C=0$
$\therefore \sqrt{x}=\frac{\alpha }{2}t$
Hence, velocity, $v=\frac{{\alpha }^2}{2}t(\because v=\alpha \sqrt{x})$
and acceleration, $a=\frac{dv}{dt}=\frac{{\alpha }^2}{2}$