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Step-by-Step Solution
Step 1: Understand the Physical Situation
The particle moves on a horizontal plane under a constant power source. Power ($P$) is the rate of doing work and can be expressed as the product of force ($F$) and velocity ($v$):
$P = F \, v$.
Step 2: Relate Power to Acceleration
If the mass of the particle is $m$, and the acceleration is $a = \frac{dv}{dt}$, we can write the force as $F = m \, \frac{dv}{dt}$. Hence,
$P = F \, v = m \, \frac{dv}{dt} \, v$.
Step 3: Separate and Integrate to Find Velocity
Rewriting,
$P = m \, v \, \frac{dv}{dt} \implies v \, dv = \frac{P}{m} \, dt.$
Integrate both sides:
$\int v \, dv = \int \frac{P}{m} \, dt.$
This gives
$\frac{v^2}{2} = \frac{P}{m} \, t + C.$
Assuming the particle starts from rest at $t = 0,$ the constant $C = 0,$ so
$v^2 = \frac{2P}{m}\, t.$
Thus,
$v = k \sqrt{t},$
where $k = \sqrt{\frac{2P}{m}}.$
Step 4: Relate Velocity to Displacement
We know $v = \frac{ds}{dt}$. Substituting $v = k \sqrt{t}$, we get
$\frac{ds}{dt} = k \sqrt{t}.$
Therefore,
$\int ds = \int k \sqrt{t} \, dt.$
Symplifying,
$s = k \int \sqrt{t} \, dt = k \cdot \frac{2}{3} \, t^{3/2} + C'.$
Again, assuming $s = 0$ at $t = 0,$ we have $C' = 0$. Hence,
$s \propto t^{3/2}.$
Step 5: Interpret the Displacement-Time Graph
A displacement that grows as $t^{3/2}$ curves upward more steeply as time increases. Among the given schematic options, the graph that shows an increasingly steep slope with time (i.e., a concave-up curve) corresponds to
.
This matches the displacement-time relationship $s \propto t^{3/2}.$
