Question

A particle located at $x=0$ at $t=0$, starts moving along the positives $x-$direction with a velocity ${}^{\prime }{v}^{\prime }$ which varies as $v=\alpha \sqrt{x}$, then velocity of particle varies with time as : ($\alpha$ is a constant)

• A. $v\propto t$
• B. $v\propto t^2$
• C. $v\propto \sqrt{t}$
• D. $v=$ constant

Answer 1

The correct option is A $v\propto t$

The velocity is given as,

$v=\alpha \sqrt{x}$

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

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

$\int \frac{dx}{\sqrt{x}}=\int \alpha dt$

$2\sqrt{x}=\alpha t+C$

Since, at $t=0$, $x=0$ so the value of $C$ is $0$.

$2\sqrt{x}=\alpha t$

$x=\frac{1}{4}{\alpha }^2{t}^2$

The velocity is given as,

$v=\frac{dx}{dt}$

$=\frac{1}{2}{\alpha }^{2}t$

Thus, the velocity of particle is directly proportional to the time $\left(v\propto t\right)$.