Problem 13 - Master Resource Book in JEE Main Physics

Question

The acceleration of a particle is increasing linearly with time \(t\) as \(b t\). The particle starts from the origin with an initial velocity \(v_{0}\). The distance travelled by the particle in time \(t\), is (a) \(v_{0} t+\frac{1}{3} b t^{2}\) (b) \(v_{0} t+\frac{1}{6} b t^{3}\) (c) \(v_{0} f+\frac{1}{3} b t^{3}\) (d) \(v_{0} t+\frac{1}{2} b t^{2}\)

Step-by-Step Solution

Answer

The correct answer is option (b): \(v_{0} t + \frac{1}{6} b t^{3}\).

Step 1: Understand Given Information

The problem provides that the acceleration \(a\) is linearly increasing with time as \(a = b t\), and the initial velocity \(v_0\) is given. We need to find the distance travelled by the particle at time \(t\).

Step 2: Acceleration to Velocity

First, recall that velocity \(v(t)\) is the integral of acceleration with respect to time. So, integrate the acceleration function \(a(t) = b t\) to find the velocity:\[ v(t) = \int b t \, dt = \frac{b t^2}{2} + C \]Here, \(C\) is the constant of integration. Since the initial velocity at \(t = 0\) is \(v_0\), we have:\[ v(0) = \frac{b (0)^2}{2} + C = v_0 \]Thus, \(C = v_0\). The velocity function becomes:\[ v(t) = \frac{b t^2}{2} + v_0 \]

Step 3: Velocity to Position

Next, integrate the velocity function \(v(t) = \frac{b t^2}{2} + v_0\) with respect to time to find the position \(x(t)\):\[ x(t) = \int \left(\frac{b t^2}{2} + v_0\right) dt = \left(\frac{b}{2} \cdot \frac{t^3}{3}\right) + v_0 t + C_1 \]where \(C_1\) is another constant of integration. Since the particle starts from the origin, \(x(0) = 0\), this gives:\[ x(0) = \frac{b}{6} (0)^3 + v_0 (0) + C_1 = 0 \]Thus, \(C_1 = 0\), and the position function simplifies to:\[ x(t) = v_0 t + \frac{b t^3}{6} \]

Step 4: Identify Correct Option

Now, compare the derived expression for \(x(t)\) with the provided options to find the correct one:1. (a) \(v_{0} t + \frac{1}{3} b t^{2}\)2. (b) \(v_{0} t + \frac{1}{6} b t^{3}\)3. (c) \(v_{0} t + \frac{1}{3} b t^{3}\)4. (d) \(v_{0} t + \frac{1}{2} b t^{2}\)The derived function \(x(t) = v_0 t + \frac{b t^3}{6}\) matches option (b).

Key Concepts

Kinematics

Kinematics is a branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. It primarily involves the study of displacement, velocity, and acceleration. In this context:

• Displacement refers to the change in position of a particle from its initial point to its final point.
• Velocity is the rate of change of displacement concerning time.
• Acceleration is the rate of change of velocity with respect to time.

In our exercise, the problem presents us with a scenario where acceleration changes with time, which is a key part of kinematics. Understanding kinematics helps in determining how particles move when subjected to certain conditions. Here, the given acceleration linearly increasing with time reflects velocity change over time, allowing us to determine the particle's position, representing its movement through space.

Linear Acceleration

Linear acceleration refers to an acceleration that changes in a linear manner with time. In simpler terms, it's when acceleration increases or decreases at a constant rate as time progresses. In our exercise, acceleration is expressed as a function of time, specifically as \( a(t) = bt \). This means that each second, the acceleration increases by a fixed amount called \( b \), the slope of the acceleration-time graph.

• This implies a non-constant, dynamically changing velocity, which updates at every moment as time passes.

Understanding linear acceleration is crucial because it helps us predict how quickly an object will speed up or slow down. This linear function displays a straightforward relationship with time, simplifying calculations when solving for velocity or position.

Integration in Physics

Integration is a powerful mathematical tool used in physics to solve problems involving changes like velocity and displacement. When acceleration is known, such as in our exercise, we can use integration to find velocity. Conversely, knowing velocity allows us to find displacement. In this problem, the acceleration function \( a(t) = bt \) was integrated to find velocity: \[ v(t) = \int bt \, dt = \frac{bt^2}{2} + C \] We use an initial condition, \( v_0 \), to solve for the constant \( C \).

• Next, integrate the velocity function to get the position function, which indicates the total distance traveled by the particle.

This layered approach of integrating step by step enables us to establish a relationship between acceleration, velocity, and position, clearly illustrating the benefits of integration in physics by connecting changes in motion with accumulated effects over time.