Question

A particle located at x = 0, time t = 0 starts moving along positive x-direction with a velocity v that varies as ${\text{v}} = \alpha \sqrt x $. The displacement of particle varies with time as
$
  {\text{A}}{\text{. }}{\alpha ^2}{t^2} \\
  {\text{B}}{\text{. }}\dfrac{{{\alpha ^2}{t^2}}}{2} \\
  {\text{C}}{\text{. }}\dfrac{{{\alpha ^2}{t^2}}}{4} \\
  {\text{D}}{\text{. None}} \\
$

Answer 1

Hint: We are given the velocity v of a particle as a function of displacement x. We know that velocity is given as the rate of change of velocity of the particle. By using this relation, we can derive the relation between displacement and time.

Formula used:
The velocity of a particle is given as
${\text{v}} = \dfrac{{dx}}{{dt}}$
where v denotes the velocity of the particle, x denotes the displacement of the particle and t is the time taken.

Detailed step by step solution:
The velocity of a particle is defined as the time rate of change of displacement of the particle. Instantaneous velocity is equal to the derivative of displacement with respect to time.
We are given a particle which is initial at x = 0 at t = 0. These are the initial coordinates of the particle.
This particle starts moving in positive x-direction with velocity v which is dependent on the displacement of the particle by the following relation.
${\text{v}} = \alpha \sqrt x $
Now we know that velocity is given as equation (i). Using this equation here we get
$
  \dfrac{{dx}}{{dt}} = \alpha \sqrt x \\
   \Rightarrow dx = \alpha \sqrt x dt \\
   \Rightarrow \dfrac{{dx}}{{\sqrt x }} = \alpha dt \\
$
Now we can perform integration on the both sides of this equation as follows:
$
  \int\limits_0^x {{x^{\dfrac{{ - 1}}{2}}}dx} = \alpha \int\limits_0^t {dt} \\
   \Rightarrow \left[ {\dfrac{{{x^{\dfrac{{ - 1}}{2} + 1}}}}{{^{\dfrac{{ - 1}}{2} + 1}}}} \right]_0^x = \alpha \left[ t \right]_0^t \\
   \Rightarrow \left[ {2{x^{\dfrac{1}{2}}} - 0} \right] = \alpha \left( {t - 0} \right) \\
   \Rightarrow 2{x^{\dfrac{1}{2}}} = \alpha t \\
$
Now we need to take square on both sides of this equation as follows:
$
  4x = {\alpha ^2}{t^2} \\
  \therefore x = \dfrac{{{\alpha ^2}{t^2}}}{4} \\
$
This is the required answer. Hence, the correct answer is option C.

Note: Instantaneous velocity of a particle is defined for a particle whose velocity is changing at every point of the trajectory. Here the velocity is a function of displacement. Since with the motion of a particle, the displacement changes at every instant, its velocity also changes at every instant because of its dependence on displacement.