Problem 12 - Master Resource Book in JEE Main Physics

Question

Power supplied to a particle of mass \(2 \mathrm{~kg}\) varies with time as \(P=\frac{3 t^{2}}{2}\) watt. Here \(t\) is in second. If the velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at time \(t=2\) s will be (a) \(1 \mathrm{~ms}^{-1}\) (b) \(4 \mathrm{~ms}^{-1}\) (c) \(2 \mathrm{~ms}^{-1}\) (d) \(2 \sqrt{2} \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

The velocity of the particle at time \(t=2\) s is \(2 \, \text{m/s}\).

Step 1: Understand the problem description

We are given that the power supplied to a particle changes with time according to the formula \(P=\frac{3t^{2}}{2}\) watts. The particle has a mass of 2 kg, and its initial velocity at \(t=0\) is 0 m/s. We need to find the velocity of the particle at \(t=2\) seconds.

Step 2: Express power in terms of work and velocity

Power \(P\) is the rate at which work is done, so \(P=\frac{dW}{dt}\). Work done \(dW\) can also be expressed as the change in kinetic energy, \(dW=\frac{d}{dt}\left(\frac{1}{2}mv^{2}\right)\). Thus, \[\frac{3t^{2}}{2} = \frac{d}{dt}\left(\frac{1}{2}mv^{2}\right).\] Given \(m=2\) kg, the equation becomes \[\frac{3t^{2}}{2} = \frac{d}{dt}(v^{2}),\] since mass \(m\) cancels out.

Step 3: Integrate with respect to time

To find \(v\) as a function of \(t\), integrate both sides of \(\frac{3t^{2}}{2} = \frac{d}{dt}(v^{2})\) with respect to \(t\):\[\int \frac{3t^{2}}{2} \, dt = \int \frac{d(v^{2})}{dt} \, dt.\]The left side integrates to \(\frac{t^{3}}{2}\), and the right side to \(v^{2}\). Therefore,\[v^{2} = \frac{t^{3}}{2} + C.\]

Step 4: Determine the constant of integration

Since the initial velocity at \(t=0\) is \(v=0\), we substitute \(t=0\) and \(v=0\) into the equation \(v^{2} = \frac{t^{3}}{2} + C\) to find \(C\). This gives\[0 = 0 + C,\]so \(C=0\). Hence, the equation simplifies to\[v^{2} = \frac{t^{3}}{2}.\]

Step 5: Find the velocity at \(t=2\) seconds

Substitute \(t=2\) into the equation \(v^{2} = \frac{t^{3}}{2}\) to determine \(v\):\[v^{2} = \frac{2^{3}}{2} = \frac{8}{2} = 4.\]Taking the square root of both sides, we find\[v = \sqrt{4} = 2 \, \text{m/s}.\]

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It is an essential concept in physics, especially when dealing with moving particles and objects. The formula for kinetic energy (KE) is given by:

• \( KE = \frac{1}{2} mv^2 \)

where \( m \) is the mass of the object, and \( v \) is its velocity. This formula tells us that the kinetic energy of an object increases with its mass and the square of its velocity.
In this exercise, we saw that the work done on a particle was equated to the change in kinetic energy using calculus. By understanding how power is related to this kinetic energy, we can delve deeper into motion mechanics.
When a force causes a change in the velocity of a particle, there is a corresponding change in kinetic energy. This change is essentially the work done on the particle, resulting in updated kinetic energy and velocity.

Integration in Calculus

Integration is a fundamental tool in calculus used to find quantities given their rates of change. In physics, it is often used to determine a function when its derivative is known.
In our exercise, we needed to find the velocity of a particle, knowing how power varies with time. The power was related to the rate of change of kinetic energy, which necessitated integrating both sides of the equation:

• \( \frac{3t^{2}}{2} = \frac{d}{dt}(v^{2}) \)
• Integrating, \( \int \frac{3t^{2}}{2} \, dt = \int \frac{d(v^{2})}{dt} \, dt \) leads to \( v^{2} = \frac{t^{3}}{2} + C \)

This process highlights how integration helps in moving from a rate of change (derivative) to the accumulated total quantity (the function itself), such as velocity from acceleration or displacement from velocity.
Integration isn't just limited to calculus problems; it allows scientists to describe real-world phenomena where quantities vary continuously over time.

Initial Conditions in Physics

Initial conditions are crucial in physics because they allow us to solve equations of motion and determine specific values from general solutions.

• When solving differential equations, like the one we encountered when relating power to kinetic energy, initial conditions help find constants of integration.

In this exercise, we knew that the velocity at time \( t=0 \) was \( 0 \, \text{m/s} \). This information let us determine the constant \( C \) in the integrated equation:

• \( v^{2} = \frac{t^{3}}{2} + C \)
• For \( t=0 \), substituting values gave \( C = 0 \)

Such conditions enable us to find the unique solution that applies to the specific scenario, which in this case, allowed us to calculate that the velocity at \( t=2 \) seconds was \( 2 \, \text{m/s} \).
Understanding and applying initial conditions ensures that our solutions are not just mathematically correct but also physically meaningful.