Q125 / 155 · WBJEE 2019 · Physics

A body starts from rest, under the action of an engine working at a constant power and moves along a straight line. The displacement s is given as a function of time ( $f$ as

  • A
    $s=a t+b t^{2}, a$ and $b$ are constants
  • B
    $s=b t^{2}, b$ is a constant
  • Correct answer
    C
    $s=a t^{3 / 2},$ a is a constant
  • D
    $s=a t,$ a is a constant

Solution

Given, Power $(P)=$ constant Kinetic Energy (KE) $=\frac{1}{2} m v^{2}$ We know that, $P=\frac{K E}{\Delta t} \Rightarrow P=\frac{m v^{2}}{\Delta t}$ $\because P=$ constant. Hence, velocity of the body $v \propto \sqrt{t}$...(i) As, Velocity $v=\frac{d s}{d t}$...(ii) From Eqs. (i) and (ii), we get So, $\quad \frac{d s}{d t} \propto \sqrt{t}$ Integrating the above cquation w.r.t. time $(t)$ $$ \int \frac{d s}{d t} \propto \int \sqrt{t} $$ we get, displacement of the body $s \propto t^{3 / 2}$ Displacement $s=a t^{3 / 2}$, where $a$ is constant.

Hence, the correct answer is option: 'C'.

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