A body is moved along a straight line by a machine delivering constant power. The distance moved by

Q: Wait, why can't we just use $s = ut + \frac{1}{2}at^2$?

That equation is strictly for **constant acceleration**. In this problem, the power $P = Fv$ is constant. As the velocity $v$ increases, the force $F$ must decrease to keep the product constant. Since force is changing, acceleration is changing, making standard kinematic equations invalid.

Q: Why do we assume the initial velocity is zero?

The problem asks for a direct proportionality of distance with time. If the body had an initial velocity $u$, the kinetic energy would be $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = Pt$. This would lead to a complex expression for $v$ that doesn't yield a simple power-law proportionality for distance. In such JEE problems, 'starts moving' or implied proportionality assumes starting from rest.

Q: How did we get $W = Pt$ without knowing the force?

Power is defined as the rate of doing work, $P = \frac{dW}{dt}$. If power is constant, we can simply multiply it by time to get the total work done, just like distance is speed times time for constant speed. We don't need to know the varying force to find the total work!

Q: Why did we use the Work-Energy theorem instead of Newton's Second Law?

You *can* use Newton's Second Law ($P = mav = m v \frac{dv}{dt}$), and it will give you the exact same result after integration! However, the Work-Energy theorem ($Pt = \frac{1}{2}mv^2$) is a powerful shortcut that skips the first integration step entirely.

Q: What happens to the acceleration as time goes on?

Since $v \propto t^{1/2}$, acceleration $a = \frac{dv}{dt} \propto t^{-1/2}$. This means as time increases, the acceleration approaches zero. The body keeps speeding up, but at a slower and slower rate!

Question

A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

Options:

(A) $t^{1/2}$

(B) $t^{3/4}$

(C) $t^{3/2}$

(D) $t^2$

Sigma Percentile · Accepted Answer

### The Power of Constant Power

Imagine you are driving a car and you floor the accelerator. Initially, you feel a massive push throwing you back into your seat. But as the car speeds up, that pushing force gradually fades, even though the engine is still working just as hard. Why does this happen?

This is the physical reality of **constant power**. Power is the rate at which work is done, or the rate at which energy is transferred. Mathematically, power is the product of force and velocity:
$$P = F \cdot v$$

If a machine delivers a constant power $P$, then as the velocity $v$ of the body increases, the force $F$ applied by the machine must decrease to keep the product constant. Because the force is constantly changing, the acceleration is also constantly changing. This is a massive trap for many students! You cannot use the standard equations of motion like $s = ut + \frac{1}{2}at^2$ because they strictly require a constant acceleration. We need a more powerful tool to solve this problem.

### The Work-Energy Connection

When dealing with varying forces but known power, the **Work-Energy Theorem** is our ultimate shortcut. It states that the net work done on an object is equal to its change in kinetic energy.

Since the power $P$ is constant, finding the total work done over a time $t$ is incredibly simple. We don't need to worry about the changing force; we just multiply the constant rate of work by the time:
$$W = \int P dt = P \cdot t$$

Assuming the body starts from rest, its initial kinetic energy is zero. Its final kinetic energy after time $t$ is $\frac{1}{2}mv^2$. Equating the work done to the change in kinetic energy gives us our master equation:
$$\frac{1}{2}mv^2 = P \cdot t$$

From this elegant relationship, we can easily isolate the velocity $v$. Rearranging the terms, we get:
$$v^2 = \frac{2P}{m} t$$

Taking the square root of both sides reveals how velocity evolves with time:
$$v = \sqrt{\frac{2P}{m}} t^{1/2}$$

Since $2$, $P$, and $m$ are all constants, we can clearly see that the velocity is directly proportional to the square root of time:
$$v \propto t^{1/2}$$

### From Velocity to Distance

We have successfully found how velocity depends on time, but the question asks for the distance moved, $s$. To bridge the gap between velocity and distance, we return to the fundamental definition of velocity as the rate of change of position:
$$v = \frac{ds}{dt}$$

Substituting our proportionality into this kinematic relation, we get a simple differential equation:
$$\frac{ds}{dt} \propto t^{1/2}$$

To find the total distance, we separate the variables, bringing the time differential $dt$ to the right side:
$$ds \propto t^{1/2} dt$$

Now, we integrate both sides. The left side integrates to the total distance $s$, and the right side requires the power rule for integration. 
$$s \propto \int t^{1/2} dt$$

Using the power rule $\int t^n dt = \frac{t^{n+1}}{n+1}$, we add $1$ to the exponent $\frac{1}{2}$ to get $\frac{3}{2}$, and divide by the new exponent:
$$s \propto \frac{t^{3/2}}{3/2}$$

Constants of proportionality absorb the numerical factor $\frac{3}{2}$, leaving us with the beautiful final result:
$$s \propto t^{3/2}$$

This tells us that under constant power, the distance covered grows with the $1.5$ power of time. It's a classic result that perfectly demonstrates the interplay between power, energy, and kinematics, completely bypassing the messy reality of a decreasing force!

Animated Solution for Physics - Work, Energy, and Power: A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

Select Answer:

Visualized Solution

Physical Setup

Definition of Work

Work Done in Time $t$

Work-Energy Theorem

Equating Work and Energy

Isolating $v^{2}$

Velocity Proportionality

Kinematic Relation

Differential Equation

Separation of Variables

Integration

Final Proportionality

The Sigma Insight: Kinetic Energy, Potential Energy and Power

The Power of Constant Power

The Work-Energy Connection

From Velocity to Distance

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