JEE PYQ: Motion in a Straight Line - Question ID 41cf457eced1 (JEE Main 2024)

Core Formula & Concept:

In kinematics, when a particle moves along a straight line, its velocity $v$ and acceleration $a$ are related to its position $x$ through calculus. The key relationships are:

• Velocity is the time derivative of position: $v = \frac{dx}{dt}$.
• Acceleration is the time derivative of velocity: $a = \frac{dv}{dt}$.

However, if velocity is given as a function of position, $v = v(x)$, we can use the chain rule to express acceleration in terms of $x$: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}.$ This formula allows us to compute acceleration without explicitly knowing time $t$.

Step-by-Step Derivation:

Given:

• Initial position: $x = 0$ at $t = 0$.
• Velocity as a function of position: $v = 4 \sqrt{x}$ m/s.

Goal: Find the acceleration $a$ of the particle.

Step 1: Express acceleration using the chain rule.

Since $v = v(x)$, we use: $a = \frac{dv}{dt} = v \cdot \frac{dv}{dx}.$

Step 2: Compute $\frac{dv}{dx}$.

Given $v = 4 \sqrt{x} = 4 x^{1/2}$, we differentiate with respect to $x$: $\frac{dv}{dx} = 4 \cdot \frac{1}{2} x^{-1/2} = 2 x^{-1/2} = \frac{2}{\sqrt{x}}.$

Step 3: Substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula.

$a = v \cdot \frac{dv}{dx} = 4 \sqrt{x} \cdot \frac{2}{\sqrt{x}}.$

Step 4: Simplify the expression.

The $\sqrt{x}$ terms cancel out: $a = 4 \cdot 2 = 8 \text{ m/s}^2.$

Conclusion:

The acceleration of the particle is constant and equal to $8 \text{ m/s}^2$. Common Traps & Exam Tip:

Students often make the following mistakes in this type of problem:

• Incorrect differentiation: Forgetting to apply the chain rule or misapplying the power rule when differentiating $v = 4 \sqrt{x}$. For example, writing $\frac{dv}{dx} = 4$ instead of $\frac{2}{\sqrt{x}}$.
• Directly differentiating $v$ with respect to $t$: Attempting to find $\frac{dv}{dt}$ without using the chain rule, which leads to confusion since $v$ is given as a function of $x$, not $t$.
• Assuming acceleration is zero or varies with $x$: Since the $\sqrt{x}$ terms cancel out, acceleration is constant. Students may overlook this simplification and conclude that acceleration depends on $x$.

Exam Tip: Always check if velocity is given as a function of position. If so, use $a = v \cdot \frac{dv}{dx}$ to find acceleration efficiently. This avoids unnecessary integration or differentiation with respect to time.