Q95 / 150 · JEE Main 2006 · Physics

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
    $t^3$
  • Correct answer
    B
    $t^2$
  • C
    $t$
  • D
    $t^{1 / 2}$

Solution

$\frac{\mathrm{d} x}{\mathrm{dt}}=\alpha \int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}=\int \alpha \mathrm{dt} \Rightarrow \mathrm{x} \alpha \mathrm{t}^2$

Hence, the correct answer is option: 'B'.

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