Problem 29 - Master Resource Book in JEE Main Physics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Given Data

We are given that the acceleration of the particle is\( a(t) = b \cdot t\). The particle starts from the origin with an initial velocity\( v_0\). We are asked to find the distance travelled by the particle in time \( t \).

Step 2: Understand the Relation Between Acceleration and Velocity

To find the velocity, integrate the acceleration function. The acceleration \( a(t) = b \cdot t \) is a function of time, so the velocity \( v(t) \) can be found using the integral of acceleration:\[ v(t) = \int a(t) \, dt = \int b \cdot t \, dt = \frac{1}{2} b t^2 + C\]The constant \( C \) is the initial velocity \( v_0 \). So,\[ v(t) = \frac{1}{2} b t^2 + v_0\]

Step 3: Calculate the Distance using Velocity

Distance or displacement is obtained by integrating the velocity function over time:\[ s(t) = \int v(t) \, dt = \int \left(\frac{1}{2} b t^2 + v_0\right) \, dt\]This leads to:\[ s(t) = \frac{1}{6} b t^3 + v_0 t + C'\]Since \( C' \) is the integration constant representing initial displacement, which is zero here (starting from the origin), we have:\[ s(t) = v_0 t + \frac{1}{6} b t^3\]

Step 4: Verify the Solution with Given Options

Now, match the obtained expression for \( s(t) \), which is \( v_0 t + \frac{1}{6} b t^3 \), with the given options. The correct solution is option (a) \( v_{0} t+\frac{1}{6} b t^{3}\).

Key Concepts

Acceleration

Acceleration is a fundamental concept in kinematics that describes how the velocity of an object changes with time. In this particular exercise, the particle's acceleration changes linearly with time, meaning that the acceleration increases steadily as time progresses. This is represented by the function \( a(t) = b \cdot t \), where \( b \) is a constant indicating how quickly the acceleration increases. Acceleration is crucial because it influences both the velocity and the displacement of the particle over time. By understanding how acceleration works, we can predict the particle's motion and determine its velocity at any point in time. When dealing with kinematics, remember that acceleration isn't fixed—it can vary, as in this example, which directly affects overall movement.

Integral Calculus

Integral calculus is a key mathematical tool used to derive other quantities from given rates of change. In this scenario, we use integration to move from acceleration to velocity and then from velocity to displacement. To find velocity from acceleration, we integrate the acceleration function \( a(t) = b \cdot t \), leading to the velocity function \( v(t) = \frac{1}{2} b t^2 + v_0 \). This process essentially accumulates the small changes in acceleration over time to find how speed evolves. Similarly, integrating the velocity function yields the displacement: \( s(t) = v_0 t + \frac{1}{6} b t^3 \). By applying integral calculus, we transition from knowing the rate of change (acceleration) to understanding the total change in position over time (displacement), making it an invaluable process in solving kinematics problems.

Displacement

Displacement refers to the change in position of a particle over a given period. In kinematics, it's important to discern displacement from distance, as displacement considers direction, while distance does not. In this exercise, displacement is determined by integrating the velocity function, resulting in \( s(t) = v_0 t + \frac{1}{6} b t^3 \). This function tells us how far the particle has traveled from its starting point in time \( t \). Note that displacement is influenced by both the initial velocity \( v_0 \) and the time-dependent acceleration. Displacement helps convey not just how far a particle has moved but also in which direction, providing a complete picture of its motion from an initial position to its final location.