A particle is moving unidirectionally on a horizontal plane under the action of a constant power sup

Q: Why can't we just use $P = Fv$ and say $F = ma$?

You absolutely can! $P = Fv = (ma)v = m \frac{dv}{dt} v$. This leads to the exact same differential equation $mv \frac{dv}{dt} = P$. Both paths are physically identical and mathematically sound.

Q: Why does the graph start with a slope of zero?

The slope of a displacement-time graph represents velocity. Since the particle starts from rest, its initial velocity is zero. Therefore, the graph must be perfectly horizontal at the origin.

Q: How do we know the graph is concave up?

The relationship is $s \propto t^{3/2}$. If we take the second derivative, $\frac{d^2s}{dt^2}$, we get acceleration. Since $v \propto t^{1/2}$, acceleration $a \propto t^{-1/2}$, which is positive for all $t > 0$. A positive second derivative means the graph is concave up.

Q: What if the power wasn't constant?

If power varied with time, say $P = kt$, you would substitute that into the integral: $\int v dv = \int \frac{kt}{m} dt$. This would change the power dependence of velocity and ultimately the shape of the displacement graph.

Q: Is it possible for the particle to have a constant acceleration here?

No. For acceleration to be constant, velocity must be proportional to $t$ ($v \propto t$). But here, $v \propto t^{1/2}$. This means the acceleration is actually decreasing over time, even though the particle is speeding up!

Question

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

Sigma Percentile · Accepted Answer

### The Power of Constant Energy

Imagine a particle of mass $m$ resting peacefully on a perfectly smooth horizontal plane. Suddenly, an engine kicks in, supplying energy to our particle at a strictly constant rate.

This means the power, $P$, delivered to the particle is constant.

To figure out how the particle's displacement changes with time, we need to build a bridge between this abstract concept of power and the tangible kinematics of motion.

**The Work-Energy Theorem** provides this exact bridge. It tells us that power is simply the rate at which kinetic energy changes over time.

### Unlocking the Kinematics

Let's write the kinetic energy as $\frac{1}{2}mv^2$.

According to our theorem, the time derivative of this kinetic energy must be equal to our constant power $P$:

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

Now, let's carefully differentiate this expression. The mass $m$ and the factor of $\frac{1}{2}$ are constants, so they pull right out of the derivative.

The derivative of $v^2$ with respect to time requires the chain rule, giving us $2v \frac{dv}{dt}$.

The twos cancel out beautifully, leaving us with a clean differential equation:

$$ mv \frac{dv}{dt} = P $$

### The First Integration: Finding Velocity

We now have a differential equation that we can solve by separating the variables.

Let's keep $v$ and $dv$ on the left side, and move $dt$ and $m$ to the right side:

$$ v \, dv = \frac{P}{m} \, dt $$

Assuming the particle starts from rest, we integrate both sides from time $0$ to $t$:

$$ \int_0^v v \, dv = \int_0^t \frac{P}{m} \, dt $$

Evaluating these integrals gives us:

$$ \frac{v^2}{2} = \frac{P}{m}t $$

Taking the square root, we find that the velocity $v$ is proportional to the square root of time:

$$ v = \sqrt{\frac{2P}{m}} t^{1/2} $$

### The Final Integration: Finding Displacement

But we aren't done yet! We are looking for displacement $s$, not velocity.

Remember the fundamental definition of velocity: it is the rate of change of displacement, $v = \frac{ds}{dt}$.

Let's substitute this into our equation:

$$ \frac{ds}{dt} = \sqrt{\frac{2P}{m}} t^{1/2} $$

We need to integrate one more time. Separating variables again:

$$ ds = \sqrt{\frac{2P}{m}} t^{1/2} \, dt $$

Integrating both sides from $t=0$ to $t$:

$$ \int_0^s ds = \int_0^t \sqrt{\frac{2P}{m}} t^{1/2} \, dt $$

By the power rule of integration, $t^{1/2}$ becomes $\frac{t^{3/2}}{3/2}$.

$$ s = \frac{2}{3}\sqrt{\frac{2P}{m}} t^{3/2} $$

### Decoding the Graph

This is our master equation! It tells us that displacement $s$ is directly proportional to $t^{3/2}$:

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

How does this look on a graph?

Because the power is $1.5$ (which is greater than $1$), the graph will curve upwards, meaning it is **concave up**.

Furthermore, if we look at the slope of this graph (which is velocity), we know that at $t=0$, the velocity is zero. Therefore, the graph must start perfectly flat at the origin.

Looking at our options, the curve that starts at the origin with a zero slope and curves upwards is exactly what is shown in **Option (a)**.

Select Answer:

Visualized Solution

$P = constant$

$Power and KE$

$Substitute K$

$Differentiate$

$Separate Variables$

$Integrate for v$

$Velocity to Displacement$

$Integrate for s$

$Final Relation$

$Graph Analysis$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Power of Constant Energy

Unlocking the Kinematics

The First Integration

The Final Integration

Decoding the Graph

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Power and KE

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute K

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differentiate

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separate Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrate for v

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity to Displacement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integrate for s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Graph Analysis