A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time

Exercises · 5.10

t

(i)

t^{1/2}

(ii)

t

(iii)

t^{3/2}

(iv)

t^{2}

Assam AhsecTextbookSubjective· 1mImportance★★★★★est

27% · 22/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Power Time Relation

Power and Time: The Intuition

Think about lifting a heavy box. If you lift it slowly, you feel tired but you manage. If you lift it fast — the same box to the same height — you feel a much greater strain. The work done (force × distance) is identical in both cases. So why does the fast lift feel harder?

The answer is power. Power tells you not just how much work is done, but how quickly it is done. The fast lift requires more power because the same work is compressed into a shorter time.

Note

Everyday language often confuses power with energy. A "powerful" car isn't one that uses more fuel (energy) — it's one that can accelerate faster or climb a hill at higher speed, meaning it delivers energy more quickly.

The Precise Statement

Power is defined as the rate at which work is done (or energy is transferred). Mathematically:

P = \frac{W}{t}

where:

$P$ = power (watts, W)
$W$ = work done (joules, J)
$t$ = time taken (seconds, s)

This is the power-time relation in its simplest form: power is work divided by time.

P = \frac{W}{t}

If you rearrange it, you get two other useful forms:

W = P \times t \quad \text{and} \quad t = \frac{W}{P}

These tell you:

To do a fixed amount of work, using more power means less time.
To run a device for a fixed time, more power means more work (and more energy consumed).

A Concrete Example

Suppose you need to lift a 10 kg mass to a height of 2 metres. The work done against gravity is:

W = mgh = 10 \times 9.8 \times 2 = 196 \text{ J}

Now consider two cases:

Case	Time taken	Power required
Slow lift	4 seconds	$P = \frac{196}{4} = 49 \text{ W}$
Fast lift	1 second	$P = \frac{196}{1} = 196 \text{ W}$

The fast lift requires four times the power — that's why it feels much harder, even though the work is the same.

Watch out

A common mistake is to think power and work are the same thing. They are not. Work is the total energy transferred; power is the rate of that transfer. A 100 W bulb uses 100 J of energy every second, but if you leave it on for an hour, the total work (energy used) is $100 \times 3600 = 360,000 \text{ J}$ .

Why This Matters for Exams

The power-time relation appears in two main forms in problems:

Direct calculation: Given work and time, find power (or any missing variable). …

Start your 14-day free trial to unlock the full solution →

Constant power means $P = Fv = \text{constant}$ . Using $F = ma$ and integrating gives $v \propto t^{1/2}$ , so displacement $s \propto t^{3/2}$ . The correct option is (iii).

The key insight here is that "constant power" is not the same as constant force. Most students instinctively assume constant force leads to $s \propto t^2$ , but power being constant changes everything. Power is the rate at which work is done, and when that rate is fixed, the force must decrease as speed increases — that coupling is what produces the unusual $t^{3/2}$ dependence.

Let’s build this from first principles.

What does constant power mean physically?

If a source delivers constant power $P$ , then the work done on the body in time $t$ is $W = Pt$ . This work goes entirely into the kinetic energy of the body (assuming no losses). So we have:

Pt = \frac{1}{2} m v^2

This single relation already tells us $v \propto t^{1/2}$ , because $P$ and $m$ are constants. But let’s verify this properly through the force approach, which is more rigorous for finding displacement.

Write the power equation in terms of force and velocity Power is $P = F v$ , where $F$ is the net force on the body. Since $P$ is constant:

F v = \text{constant}

Replace force using Newton’s second law $F = m a = m \frac{dv}{dt}$ . Substituting:

m \frac{dv}{dt} \cdot v = P

This is a differential equation in $v$ :

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

Separate variables and integrate

\int m v \, dv = \int P \, dt

\frac{1}{2} m v^2 = P t + C

Assuming the body starts from rest ( $v=0$ at $t=0$ ), the constant $C=0$ . So:

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

This confirms $v \propto t^{1/2}$ .

Watch out

A common mistake is to treat $F$ as constant when power is constant. If $P = Fv$ is fixed, then as $v$ increases, $F$ must decrease. Constant power does not mean constant acceleration — in fact, acceleration decreases with time here.

Find displacement by integrating velocity

s = \int v \, dt = \sqrt{\frac{2P}{m}} \int t^{1/2} \, dt

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

Start your 14-day free trial to unlock the full solution →

←Previous question5.9 — A body is initially at rest. It undergoes one-dimensional motion with …→Next question5.11 — A body constrained to move along the z-axis of a coordinate system is …

Q.A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to

Concept understanding — Power Time Relation

Power and Time: The Intuition

The Precise Statement

A Concrete Example

Why This Matters for Exams

Concept: Constant Power Kinematics

Q.A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time ttt is proportional to

🎓Unlock everything free for 14 days

Q.A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to

Unlock everything free for 14 days