JEE PYQ: Motion in a Straight Line - Question ID d9ac0a553e7c (JEE Main 2024)

Core Formula & Concept:

To solve this problem, we rely on the fundamental definitions of kinematics in one dimension:

• Displacement: $x(t)$ is the position of the particle at time $t$.
• Velocity: $v = \frac{dx}{dt}$, the first derivative of displacement with respect to time.
• Acceleration: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$, the second derivative of displacement with respect to time.

Given the relation $x^2 = 1 + t^2$, we will differentiate it implicitly to find velocity and acceleration. The goal is to express acceleration in the form $x^{-n}$ and determine the value of $n$.

Step-by-Step Derivation:

Step 1: Express displacement explicitly

Given: $x^2 = 1 + t^2$ Taking the positive root (assuming $x > 0$ for simplicity, as displacement is a scalar here): $x = \sqrt{1 + t^2}$

Step 2: Find velocity by differentiating displacement

Differentiate $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt} \left( (1 + t^2)^{1/2} \right) = \frac{1}{2}(1 + t^2)^{-1/2} \cdot 2t = \frac{t}{\sqrt{1 + t^2}} = \frac{t}{x}$ Since $x = \sqrt{1 + t^2}$, we have: $v = \frac{t}{x}$

Step 3: Find acceleration by differentiating velocity

Acceleration is: $a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{t}{x} \right)$ Use the quotient rule: $a = \frac{x \cdot \frac{dt}{dt} - t \cdot \frac{dx}{dt}}{x^2} = \frac{x \cdot 1 - t \cdot v}{x^2}$ But $v = \frac{t}{x}$, so: $a = \frac{x - t \cdot \frac{t}{x}}{x^2} = \frac{x - \frac{t^2}{x}}{x^2} = \frac{\frac{x^2 - t^2}{x}}{x^2} = \frac{x^2 - t^2}{x^3}$

Step 4: Use the original relation to simplify

From $x^2 = 1 + t^2$, we get $x^2 - t^2 = 1$. Substitute this into the expression for $a$: $a = \frac{1}{x^3} = x^{-3}$

Step 5: Compare with the given form of acceleration

The problem states that acceleration is $x^{-n}$. From the above, we have: $a = x^{-3}$ Thus, $n = 3$.

Common Traps & Exam Tip:

Trap 1: Incorrect differentiation
Students often make mistakes in applying the chain rule or quotient rule when differentiating $x = \sqrt{1 + t^2}$ or $v = \frac{t}{x}$. Always double-check derivatives.

Trap 2: Forgetting to substitute $x^2 - t^2 = 1$
Without using the original relation $x^2 = 1 + t^2$, the expression for acceleration remains complicated. Always look for opportunities to simplify using given equations.

Trap 3: Sign errors in displacement
Assuming $x > 0$ is safe here, but in problems involving direction, sign errors can lead to incorrect results. Be mindful of the physical context.

Exam Tip:
When acceleration is expressed in terms of displacement, always try to eliminate time $t$ using the given relation. This often simplifies the expression and reveals the required exponent.

Final Answer: $n = \boxed{3}$