Question

A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement $\left(s\right)$ – time $\left(t\right)$ graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale):

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Step 1. Given data

A particle is moving uni-directionally on a horizontal plane under the action of a constant power supplying energy source.

Step 2. Finding the displacement-time graph

Power, $P$ is the rate of change of kinetic energy, $K.E.$ with respect to the time, $t$.

Kinetic energy is the half of the product of the mass, $m$ and the square of the velocity, $v$

As power is constant,

$P=\frac{d}{dt}\left(K.E.\right)=$constant

$P=\frac{d}{dt}\left(\frac{1}{2}m{v}^2\right)=$constant

$\frac{1}{2}\left(2mv\right)\frac{dv}{dt}=P$

$vdv=\frac{P}{m}dt$

Integrating both the sides, we get

$$\begin{gathered} \int vdv=\frac{P}{m}\int dt \\ \frac{v^2}{2}=\frac{P}{m}t \\ v=\sqrt{\frac{2Pt}{m}} \end{gathered}$$

$$\begin{gathered} v=\frac{ds}{dt} \\ \frac{ds}{dt}=\sqrt{\frac{2Pt}{m}} \\ ds=\sqrt{\frac{2Pt}{m}}dt \end{gathered}$$ [Velocity is the rate of the change of the displacement, $s$ with respect to the time, $t$

Again, integrating both the sides, we get

$$\begin{gathered} \int ds=\sqrt{\frac{2P}{m}}\int t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dt \\ s=\frac{2}{3}\sqrt{\frac{2P}{m}}{t}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Hence, option $\left(B\right)$ is correct.