Question

A particle located at x = 0 time t =0 starts moving along positive x-direction with a velocity v that varies as $v=\alpha \sqrt{x}$ displacement of particle at time t.

• A. ${\alpha }^2{t}^2$
• B. $\frac{{\alpha }^2{t}^2}{2}$
• C. $\frac{{\alpha }^2{t}^2}{4}$
• D. None

Answer 1

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

$V=\alpha \sqrt{x}$ But $V=\frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt}=\alpha \sqrt{x}$ & $\frac{dx}{\sqrt{x}}=\alpha$ dt intergrating $\frac{x^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}=\alpha t+c$ & $2\sqrt{x}=\alpha t+c$

$\Rightarrow$ at t=0 xiso so putting $2\times 0=0+c\Rightarrow c=0$

$\Rightarrow 2\sqrt{x}=\alpha t$ & $x={\alpha }^2{t}^2/4$

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