Question

A particle is moving with speed $v=b\sqrt{x}$ along positive $x-$axis. Calculate the speed of the particle at time $t=\tau$ (assume that the particle is at origin at $t=0$)

• A. $\frac{b^2\tau }{4}$
• B. $\frac{b^2\tau }{2}$
• C. $b^2\tau$
• D. $\frac{b^2\tau }{\sqrt{2}}$

Answer 1

The correct option is B $\frac{b^2\tau }{2}$

Given,

$v=b\sqrt{x}$

$\Rightarrow a=\frac{dv}{dt}=\frac{b}{2\sqrt{x}}\frac{dx}{dt}$

$\Rightarrow a=\frac{bv}{2\sqrt{x}}=\frac{b\times b\sqrt{x}}{2\sqrt{x}}[\because \frac{dx}{dt}=v]$

$\Rightarrow a=\frac{b^2}{2}$

As, $a=\frac{dv}{dt}$

$\Rightarrow \frac{dv}{dt}=\frac{b^2}{2}$

Rearranging and integrating the above equation,

${\int }_0^{v}dv={\int }_0^{\tau }\frac{b^2}{2}dt$

$\therefore v=\frac{b^2}{2}\tau$

Hence, $\text{(B)}$ is the correct answer.