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 $t$.
$\because \frac{d x}{d t}=\alpha \sqrt{x}$
$\therefore \frac{d x}{\sqrt{x}}=\alpha d t$
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, $a=\frac{d v}{d t}=\frac{\alpha^{2}}{2}$