Animated Solution for Physics - Work, Energy, and Power: An automobile of mass $m$ accelerates starting from origin and initially at rest, while the engine supplies constant power $P$. The position is given as a function of time by

When we first learn kinematics, we are heavily trained to use the classic equations of motion like $s=ut+\frac{1}{2}a{t}^2$. These equations are incredibly powerful, but they come with a strict, non-negotiable condition: the acceleration must be constant.

In the real world, however, things rarely accelerate constantly. Imagine you are driving a car. When you press the gas pedal to the floor, the engine doesn't deliver a constant force; it delivers a constant power. This is a profound difference. Power is the rate at which work is done, and as we will see, a constant power output leads to a fascinating, non-linear relationship between position and time.

Let's break down the problem. We have an automobile of mass $m$ starting from rest ($u=0$). The engine supplies a constant power $P$. We need to find its position $s$ as a function of time $t$.

Since we are dealing with power, time, and changing speeds, the Work-Energy Theorem is our master key. The theorem states that the net work done on an object equals its change in kinetic energy:

Because the power $P$ is constant, the total work done by the engine over a time interval $t$ is simply the power multiplied by the time:

Now, let's look at the kinetic energy. The car starts from rest, so its initial kinetic energy is zero. If its velocity at time $t$ is $v$, its final kinetic energy is $\frac{1}{2}m{v}^2$. Substituting these into our Work-Energy equation gives us our raw setup:

From this elegant equation, we can easily isolate the velocity $v$. Multiplying both sides by 2 and dividing by $m$, we get:

Taking the square root of both sides yields the velocity as a function of time:

Look closely at this result. The velocity is proportional to the square root of time ($v\propto \sqrt{t}$). If the acceleration were constant, velocity would be directly proportional to time ($v\propto t$). Because the velocity grows slower than a linear rate, it implies that the acceleration is actually decreasing over time.

Physically, this makes perfect sense. Power is force times velocity ($P=Fv$). If $P$ is constant, as the car speeds up and $v$ increases, the force $F$ must decrease. Less force means less acceleration.

From Velocity to Position

We have the velocity, but the question asks for the position $s$. To bridge this gap, we must return to the fundamental calculus definition of velocity. Velocity is the rate of change of position:

Substituting our expression for $v$ into this definition gives us a differential equation:

To find the total displacement $s$, we separate the variables, moving $dt$ to the right side:

Now, we integrate both sides. The car starts at the origin ($s=0$) at time $t=0$:

The integral of $ds$ is simply $s$. The integral of $t^{1/2}$ requires the power rule: add 1 to the exponent and divide by the new exponent. So, $t^{1/2}$ becomes $\frac{t^{3/2}}{3/2}$, which simplifies to $\frac{2}{3}{t}^{3/2}$.

We have our answer, but it doesn't quite look like the options provided in the question. We need to do a little algebraic housekeeping. Let's bring the fraction $\frac{2}{3}$ inside the square root. To do this, we must square it, making it $\frac{4}{9}$:

Multiplying the terms inside the parentheses:

And there we have it! This matches option (d) perfectly.

This problem is a beautiful demonstration of why we must always check our assumptions before applying formulas. The kinematic equations are a special case, not a universal law. By relying on the fundamental principles of Work and Energy, and applying a little bit of calculus, we can model complex, real-world scenarios like a car engine delivering constant power. The result, $s\propto t^{3/2}$, is a unique mathematical signature of this specific physical reality.