Problem 20 - Master Resource Book in JEE Main Physics

Question

If the velocity \(v\) of a particle moving along a straight line decreases linearly with its displacement \(s\) from \(20 \mathrm{~ms}^{-1}\) to a value approaching zero at \(s=30 \mathrm{~m}\), then acceleration of the particle at \(s=15 \mathrm{~m}\) is (a) \(\frac{2}{3} \mathrm{~ms}^{-2}\) (b) \(-\frac{2}{3} \mathrm{~ms}^{-2}\) (c) \(\frac{20}{3} \mathrm{~ms}^{-2}\) (d) \(-\frac{20}{3} \mathrm{~ms}^{-2}\)

Step-by-Step Solution

Answer

(d) \(-\frac{20}{3} \, \text{ms}^{-2}\)

Step 1: Understand the Problem

We know that the velocity of a particle decreases linearly from \(20 \, \text{ms}^{-1}\) to \(0\) at \(s = 30 \, \text{m}\). We need to find the acceleration at \(s = 15 \, \text{m}\). Acceleration is the derivative of velocity with respect to time, but since velocity is expressed in terms of displacement, we will use the chain rule.

Step 2: Express Velocity as a Function of Displacement

Since the velocity decreases linearly, we can express it in the form \(v = ms + c\). Using the values given, we know \(v = 20 \, \text{ms}^{-1}\) when \(s = 0\) and \(v = 0\) when \(s = 30 \, \text{m}\). Substitute these into the equation to find \(m\) and \(c\).

Step 3: Find the Slope and Equation of the Line

For a line, the slope \(m\) is computed as \(\frac{\Delta v}{\Delta s} = \frac{0 - 20}{30 - 0} = \frac{-20}{30} = -\frac{2}{3}\). Therefore, the equation of the line is \(v = -\frac{2}{3}s + 20\).

Step 4: Calculate the Derivative to Find Acceleration

To find acceleration \(a = \frac{dv}{dt}\), we use the chain rule: \(a = \frac{dv}{ds} \times \frac{ds}{dt}\). But since \(v = \frac{ds}{dt}\), we have: \(a = \frac{dv}{ds} \cdot v\). The derivative \(\frac{dv}{ds} = -\frac{2}{3}\).

Step 5: Determine Acceleration at Specific Displacement

Plug in \(v = -\frac{2}{3}s + 20\) into the expression for acceleration: \(a = \left(-\frac{2}{3}\right) v\). At \(s = 15\), \(v = -\frac{2}{3} \times 15 + 20 = 10\). Therefore, \(a = \left(-\frac{2}{3}\right) \times 10 = -\frac{20}{3} \, \text{ms}^{-2}\).

Step 6: Choose the Correct Answer

Now we compare our result with the given options: (a) \(\frac{2}{3} \, \text{ms}^{-2}\), (b) \(-\frac{2}{3} \, \text{ms}^{-2}\), (c) \(\frac{20}{3} \, \text{ms}^{-2}\), and (d) \(-\frac{20}{3} \, \text{ms}^{-2}\). The correct answer is option (d), \(-\frac{20}{3} \, \text{ms}^{-2}\).

Key Concepts

Linear Motion

Linear motion refers to the movement of an object in a straight line with uniform displacement over time. When we assess linear motion, important parameters are velocity, displacement, and time. In scenarios where an object’s velocity changes at a constant rate as it moves along a straight line, this is characterized as linear dependence, such as when a particle’s speed decreases linearly with increase in displacement.
For example, in the given exercise, the velocity decreases uniformly from 20 m/s to 0 m/s over a linear path of 30 meters. This indicates that the particle has linear motion with a predictable pattern. Linear relationships are beneficial because they allow us to create linear equations for precise calculations. Thus, understanding these helps immensely in predicting and analyzing an object's motion.

Velocity and Displacement

Velocity is defined as the rate at which an object changes its position. It is a vector quantity, which means it has both magnitude and direction. In the context of displacement, velocity plays a crucial role in determining how the position of an object changes over time. Displacement, on the other hand, is a vector that specifies the change in an object's position. It only considers the starting and ending points, disregarding the path taken.
In the context of this exercise, velocity decreases linearly with displacement. This means we can establish a linear relationship, such as the equation derived, where velocity is expressed as a function of displacement: \[ v = -\frac{2}{3}s + 20 \]This indicates that for every increase of 1 meter in displacement, the velocity decreases by \(-\frac{2}{3}\) m/s.

Acceleration Derivation

Acceleration is the rate of change of velocity with respect to time. In the case of velocity being a function of displacement rather than time, we can use the chain rule to derive acceleration. This involves a two-step approach: first, finding the derivative of velocity with respect to displacement, \( \frac{dv}{ds} \), and then multiplying it by velocity, \( v \), because \( a = \frac{dv}{ds} \times v \).
In the exercise scenario, since velocity changes linearly with displacement, \( \frac{dv}{ds} \) is constant at \(-\frac{2}{3}\). This constant rate of change is helpful because it defines how quickly the particle slows down as it travels along the displacement. Therefore, calculating acceleration becomes straightforward: substituting our values into \( a = \left(-\frac{2}{3} \right) \times v\), we find that at midpoint \( s = 15 \text{ m} \), the velocity \( v = 10 \text{ m/s} \). Subsequently, the acceleration \( a \) ends up being \(-\frac{20}{3} \text{ m/s}^2\), indicating a consistent deceleration as the particle progresses in its path.