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. $t^{1/2}$
• D. $t^{3}$

Answer 1

Answer: (A)

$t^{2}$

Answer 2

$v=\alpha\sqrt{x}$ $\frac{d x}{d t}=\alpha\sqrt{x} \, \quad\left(\because v=\frac{d x}{d t}\right)$ $\frac{d x}{\sqrt{x}}=\alpha \, dt$ Perform integration $\int_{0}^{x} \frac{dx}{\sqrt{x}}=\int_{0}^{1} \alpha dt$ $\left[\because at\, t = 0, \,x = 0 and \,let\, at \,any \,time t,\, particle\, is \, at \, x\right]$ $\Rightarrow \quad \frac{x^{1 /2}}{1 /2}|_{0}^{x}=\alpha t$ $\Rightarrow \quad x^{1 / 2}=\frac{\alpha}{2} t$ $\Rightarrow \quad\quad x=\frac{\alpha^{2}}{4}\times t^{2} \Rightarrow x \propto t^{2}$