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. $t^{\frac{1}{2}}$
• D. $t^3$

Answer 1

The correct option is A

$t^2$

Step 1. Given data

Velocity, $v=\alpha \sqrt{x}$

Step 2. Finding the displacement, $x$

Velocity, $v$ is the rate of change of the displacement, $x$ with respect to the time, $t$

$$\begin{gathered} v=\alpha \sqrt{x} \\ \frac{dx}{dt}=\alpha \sqrt{x} \\ \frac{1}{\sqrt{x}}dx=\alpha dt \\ {\int }_0^x\frac{1}{\sqrt{x}}dx={\int }_0^t\alpha dt \\ 2\sqrt{x}=\alpha t \\ x=\frac{{\alpha }^2}{4}{t}^2 \\ x\propto t^2 \end{gathered}$$ [ By integrating]

Hence, option $\left(A\right)$ is correct.