Question

Velocity of a particle moving in $x-$axis is given as $v=\alpha \sqrt{x}$ where $\alpha$ is positive constant. If initially particle was at origin, the position of particle at time $t$ is

• A. $\frac{{\alpha }^2{t}^2}{4}$
• B. $3{\alpha }^2\frac{t^2}{4}$
• C. ${\alpha }^2\frac{t^2}{2}$
• D. $3{\alpha }^2\frac{t^2}{2}$

Answer 1

The correct option is A $\frac{{\alpha }^2{t}^2}{4}$

$\begin{array}{c}v=\alpha \sqrt{x}\\ \Rightarrow \frac{dx}{dt}=\alpha \sqrt{x}\\ \Rightarrow \frac{dx}{\sqrt{x}}=\alpha dt\\ {\Rightarrow }^{\frac{dx}{\sqrt{x}}}={\alpha }^{d}t\\ \Rightarrow \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\alpha t+c\end{array}$

Since,at $t=0,x=0$

we get $c=0$

So,

$\begin{array}{c}2\sqrt{x}=\alpha t\\ \Rightarrow 4x={\alpha }^2{t}^2\\ \Rightarrow x=\frac{1}{4}{\alpha }^2\cdot t^2\end{array}$

Hence, Option $A$ is the correct answer.