1. Given condition: constant power

If a constant power source is supplying energy to a particle, then

$P = \frac{d}{dt}\left(\frac{1}{2}mv^2\right).$

Since $P$ is constant,

$\frac{d}{dt}\left(\frac{1}{2}mv^2\right)=P.$

Integrating,

$\frac{1}{2}mv^2 = Pt + C.$

If the particle starts from rest at $t=0$, then $C=0$, so

$v^2 = \frac{2Pt}{m}$

or

$v = \frac{ds}{dt} \propto \sqrt{t}.$

2. Find displacement as a function of time

Since

$\frac{ds}{dt} = k\sqrt{t},$

where $k$ is a constant, integrate:

$s = \int k\sqrt{t}\,dt = k\cdot \frac{2}{3}t^{3/2} + C'.$

Taking $s=0$ at $t=0$, we get

$s \propto t^{3/2}.$

3. Nature of the $s$-$t$ graph

For

$s \propto t^{3/2},$

we have:

• slope $\dfrac{ds}{dt} \propto \sqrt{t}$, so slope increases with time;
• second derivative $\frac{d^2s}{dt^2} \propto \frac{1}{\sqrt{t}} > 0,$ so the graph is concave upward;
• at $t=0$, the slope is zero.

So the correct schematic graph must:

• start with nearly zero slope,
• become steeper with time,
• be curved upward.

4. Match with options

Among the given schematic options, this corresponds to Option B.

5. Comparison with stored answer

Derived answer: B
Stored correct answer: B

They match.