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 - -x-Where -a- is a constant- Then The displacement of the particle varies with time as

The correct option is B t^2 v = α√(x) dxdt = α√(x) ∫_x = 0^xdx√(x) = α∫_t = 0^t dt ⇒ x = α^24 t^2 ⇒∝ t^2 ...

Motion Under Variable Acceleration

A particle lo...

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 $v = α \sqrt{x}$ .Where $^{'} a^{'}$ is a 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 $t^{2}$
$v = α \sqrt{x}$
$\frac{d x}{d t} = α \sqrt{x}$
$\int_{x = 0}^{x} \frac{d x}{\sqrt{x}} = α \int_{t = 0}^{t} d t$
$\Rightarrow x = \frac{α^{2}}{4} t^{2}$
$\Rightarrow\propto t^{2}$

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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 as v=α√x.Where ′a′ is a constant. Then The displacement of the particle varies with time as

The correct option is B t2v=α√xdxdt=α√x∫xx=0dx√x=α∫tt=0dt⇒x=α24t2⇒∝t2

A particle located at $x = 0$ at time $t = 0$ , starts moving along the positive $x -$ direction with a velocity $^{'} v^{'}$ that varies as $v = α \sqrt{x}$ .Where $^{'} a^{'}$ is a constant. Then The displacement of the particle varies with time as

The correct option is B $t^{2}$
$v = α \sqrt{x}$
$\frac{d x}{d t} = α \sqrt{x}$
$\int_{x = 0}^{x} \frac{d x}{\sqrt{x}} = α \int_{t = 0}^{t} d t$
$\Rightarrow x = \frac{α^{2}}{4} t^{2}$
$\Rightarrow\propto t^{2}$