JEE PYQ: Motion in a Straight Line - Question ID 9f389321e511 (JEE Main 2022)

Core Formula & Concept:

When a particle moves in a straight line, its acceleration is defined as the rate of change of velocity with respect to time: $a = \frac{dv}{dt}$ However, in this problem, the velocity is given to be increasing at a rate of $5 \text{ ms}^{-1}$ per meter. This implies that the change in velocity depends on the displacement ($s$), not directly on time. To relate these quantities, we use the chain rule of calculus: $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \cdot \frac{dv}{ds}$ Here:

• $v = \frac{ds}{dt}$ is the velocity of the particle.
• $\frac{dv}{ds}$ represents the rate of change of velocity with respect to displacement, which is given as $5 \text{ ms}^{-1}$ per meter.

Thus, the acceleration can be expressed in terms of $v$ and $\frac{dv}{ds}$.

Step-by-Step Derivation:

Given:

• The velocity increases at a rate of $5 \text{ ms}^{-1}$ per meter, i.e., $\frac{dv}{ds} = 5 \text{ m}^{-1}\text{s}^{-1}$.
• The velocity at the point of interest is $v = 20 \text{ ms}^{-1}$.

Using the chain rule expression for acceleration: $a = v \cdot \frac{dv}{ds}$ Substitute the given values: $a = 20 \text{ ms}^{-1} \cdot 5 \text{ m}^{-1}\text{s}^{-1}$ Simplify the units:

• $\text{ms}^{-1} \cdot \text{m}^{-1}\text{s}^{-1} = \text{m} \cdot \text{s}^{-2}$ (since $\text{s}^{-1} \cdot \text{s}^{-1} = \text{s}^{-2}$ and $\text{m}^{-1} \cdot \text{m} = 1$).

Thus: $a = 20 \times 5 = 100 \text{ ms}^{-2}$ The acceleration of the particle at the given point is $100 \text{ ms}^{-2}$.

Common Traps & Exam Tip:

1. Misinterpreting the given rate: Students often confuse $\frac{dv}{ds}$ with $\frac{dv}{dt}$. The problem states that velocity increases "per meter," not "per second," so $\frac{dv}{ds}$ is the correct quantity to use. Always read the units carefully.
2. Incorrect application of the chain rule: Some students forget to multiply by $v$ when converting $\frac{dv}{ds}$ to $\frac{dv}{dt}$. Remember that $a = \frac{dv}{dt} = v \cdot \frac{dv}{ds}$ is the key relationship here.
3. Unit confusion: Mixing up the units of $\frac{dv}{ds}$ (which is $\text{m}^{-1}\text{s}^{-1}$) with those of acceleration ($\text{ms}^{-2}$) can lead to errors. Always verify the units at each step to ensure consistency.
4. Assuming constant acceleration: The problem does not state that acceleration is constant. The given rate ($\frac{dv}{ds}$) may vary, but we are only asked for the acceleration at a specific velocity. Avoid unnecessary assumptions.

Exam Tip: When dealing with rates involving displacement, always think of the chain rule to relate $\frac{dv}{dt}$, $\frac{dv}{ds}$, and $\frac{ds}{dt}$. This is a common theme in kinematics problems involving non-uniform acceleration.