JEE PYQ: Motion in a Straight Line - Question ID 28f31a58bfae (JEE Main 2024)

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

• Position: Given as a function of time, $x(t)$, in meters.
• Velocity: The first derivative of position with respect to time: $v(t) = \frac{dx}{dt}$
• Acceleration: The first derivative of velocity with respect to time, or equivalently, the second derivative of position: $a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$

The question asks for the velocity of the particle at the instant when its acceleration becomes zero. This means we must:

1. Find the expression for acceleration $a(t)$ by differentiating $x(t)$ twice.
2. Set $a(t) = 0$ and solve for time $t$.
3. Substitute this time back into the velocity expression $v(t)$ to find the required velocity.

Step 1: Write the given position function

The position of the particle as a function of time is: $x(t) = -3 t^3 + 18 t^2 + 16 t$ where $x$ is in meters and $t$ is in seconds.

Step 2: Find the velocity function $v(t)$

Velocity is the first derivative of position with respect to time: $v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( -3 t^3 + 18 t^2 + 16 t \right)$ Differentiating term by term: $v(t) = -3 \cdot 3 t^{2} + 18 \cdot 2 t + 16 = -9 t^2 + 36 t + 16$ So, $v(t) = -9 t^2 + 36 t + 16$

Step 3: Find the acceleration function $a(t)$

Acceleration is the first derivative of velocity with respect to time: $a(t) = \frac{dv}{dt} = \frac{d}{dt} \left( -9 t^2 + 36 t + 16 \right)$ Differentiating: $a(t) = -9 \cdot 2 t + 36 = -18 t + 36$ So, $a(t) = -18 t + 36$

Step 4: Find the time $t$ when acceleration is zero

Set $a(t) = 0$: $-18 t + 36 = 0 \Rightarrow -18 t = -36 \Rightarrow t = \frac{36}{18} = 2 \text{ s}$ So, acceleration becomes zero at $t = 2$ seconds.

Step 5: Find the velocity at $t = 2$ s

Substitute $t = 2$ into the velocity expression: $v(2) = -9 (2)^2 + 36 (2) + 16 = -9 \cdot 4 + 72 + 16 = -36 + 72 + 16 = (72 - 36) + 16 = 36 + 16 = 52 \text{ m/s}$

Conclusion:

The velocity of the particle when its acceleration becomes zero is $52 \text{ m/s}$.

Students often make the following mistakes in such problems:

• Incorrect differentiation: Forgetting to apply the power rule correctly, especially with negative coefficients or higher powers of $t$. For example, differentiating $-3 t^3$ as $-9 t^2$ instead of $-9 t^2$ is correct, but misapplying signs or exponents can lead to errors.
• Confusing velocity and acceleration: Some students stop at finding velocity and forget to compute acceleration or vice versa. Always read the question carefully: it asks for velocity when acceleration is zero, not acceleration itself.
• Arithmetic errors: Simple addition or multiplication mistakes, especially under exam pressure, can lead to wrong final values. Double-check calculations like $-36 + 72 + 16$.
• Ignoring units: While units are given, some students forget to include them in the final answer. Always specify units (here, m/s).

Exam Tip: When dealing with polynomial position functions, always:

• Differentiate carefully, term by term.
• Set the second derivative (acceleration) to zero to find critical time.
• Substitute back into the first derivative (velocity) to find the required value.

This systematic approach ensures accuracy and builds confidence in kinematics problems.