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

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 = $α$ $\sqrt{x}$ .The displacement of the particle varies with time as

Q. A particle located at

x = 0

t = 0

, starts moving along the positives

x -

direction with a velocity

^{'} v^{'}

which varies as

v = α \sqrt{x}

, then velocity of particle varies with time as : (

α

is a constant)

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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 $t^{2}$
$v = α \sqrt{x}$
$\frac{d x}{d t} = α \sqrt{x}$
$\int_{x = 0}^{x} \frac{d x}{\sqrt{x}} = α \int_{t = 0}^{t} d t$
$\Rightarrow x = \frac{α^{2}}{4} t^{2}$
$\Rightarrow\propto t^{2}$