JEE PYQ: Motion in a Straight Line - Question ID d0d7f5cc86af (JEE Main 2021)

In kinematics, acceleration ($a$) is the rate of change of velocity ($\upsilon$) with respect to time ($t$). However, when velocity is given as a function of displacement ($x$), we use the chain rule of differentiation to express acceleration in terms of $x$:

Here, $\frac{dx}{dt} = \upsilon$ (velocity), and $\frac{d\upsilon}{dx}$ is the derivative of velocity with respect to displacement. This formula allows us to compute acceleration directly from the given velocity-displacement relationship.

Given the velocity-displacement relation:

1. Express $\upsilon$ in a differentiable form: $\upsilon = (5000 + 24x)^{1/2}.$
2. Compute $\frac{d\upsilon}{dx}$ using the chain rule: $\frac{d\upsilon}{dx} = \frac{1}{2}(5000 + 24x)^{-1/2} \cdot 24 = \frac{12}{\sqrt{5000 + 24x}}.$
3. Substitute $\upsilon$ and $\frac{d\upsilon}{dx}$ into the acceleration formula: $a = \upsilon \cdot \frac{d\upsilon}{dx} = \sqrt{5000 + 24x} \cdot \frac{12}{\sqrt{5000 + 24x}}.$
4. Simplify the expression: The $\sqrt{5000 + 24x}$ terms cancel out, leaving: $a = 12 \text{ m/s}^2.$

1. Misapplying the chain rule: Students often forget to multiply by $\upsilon$ when using $a = \upsilon \cdot \frac{d\upsilon}{dx}$, leading to incorrect results like $a = \frac{d\upsilon}{dx}$.
2. Algebraic errors in differentiation: Incorrectly computing $\frac{d\upsilon}{dx}$ (e.g., forgetting the exponent or the chain rule) can yield wrong values.
3. Unit confusion: Ensure all terms are in consistent units (here, $x$ is in meters, $\upsilon$ in m/s, and $a$ in m/s²).
4. Overcomplicating the problem: The given velocity-displacement relation simplifies neatly, so avoid unnecessary substitutions or integrations.

Exam Tip: Always verify that the units of the final answer match the expected dimensions (here, m/s² for acceleration). The cancellation of terms in this problem is a hint that the acceleration is constant.