A particle located at $x = 0$ at time $t = 0$ , starts moving along the positive $x -$ direction with a velocity $v$ that varies with position as $v = a \sqrt{x}$ . Where $a$ is constant. Then the displacement of the particle varies with time as:

t^{3}

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t^{2}

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t

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t^{1 / 2}

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Solution

The correct option is B $t^{2}$
$v = \frac{d x}{d t} = a \sqrt{x}$
$\int \frac{d x}{\sqrt{x}} = \int a d t$
$2 \sqrt{x} = a t$
$4 x = a^{2} t^{2}$
hence the answer is $t^{2}$

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A particle located at $x = 0$ at time $t = 0$ , starts moving along the positive $x -$ direction with a velocity $v$ that varies with position as $v = a \sqrt{x}$ . Where $a$ is constant. Then the displacement of the particle varies with time as:

The correct option is B $t^{2}$
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$\int \frac{d x}{\sqrt{x}} = \int a d t$
$2 \sqrt{x} = a t$
$4 x = a^{2} t^{2}$
hence the answer is $t^{2}$