JEE PYQ: Motion in a Straight Line - Question ID 25cab711879a (JEE Main 2024)

Core Formula & Concept:

In kinematics, the motion of a particle along a straight line is described by its position \( x(t) \) as a function of time \( t \). The key relationships are:

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

The problem asks for the velocity at the instant when the acceleration becomes zero. This requires:

1. Differentiating the given position function twice to obtain acceleration.
2. Setting the acceleration expression equal to zero and solving for \( t \).
3. Substituting this time back into the velocity expression to find the required velocity.

Step-by-Step Derivation:

Given the position function: $x(t) = t^3 - 6t^2 + 20t + 15$

Step 1: Compute the velocity \( v(t) \)

Differentiate \( x(t) \) with respect to \( t \): $v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 20t + 15)$ $v(t) = 3t^2 - 12t + 20$

Step 2: Compute the acceleration \( a(t) \)

Differentiate \( v(t) \) with respect to \( t \): $a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20)$ $a(t) = 6t - 12$

Step 3: Find the time \( t \) when acceleration is zero

Set \( a(t) = 0 \): $6t - 12 = 0$ $6t = 12$ $t = 2 \text{ s}$

Step 4: Compute the velocity at \( t = 2 \) s

Substitute \( t = 2 \) into \( v(t) \): $v(2) = 3(2)^2 - 12(2) + 20$ $v(2) = 3(4) - 24 + 20$ $v(2) = 12 - 24 + 20$ $v(2) = 8 \text{ m/s}$

Conclusion:

The velocity of the particle when its acceleration becomes zero is \( 8 \text{ m/s} \), which corresponds to option C.

Common Traps & Exam Tip:

Common Mistakes:

• Incorrect differentiation: Students sometimes forget to apply the power rule correctly, especially for terms like \( t^3 \) or \( t^2 \). For example, differentiating \( t^3 \) as \( 3t \) instead of \( 3t^2 \).
• Misinterpreting the question: Some students compute the velocity at \( t = 0 \) or another arbitrary time instead of finding the time when acceleration is zero.
• Sign errors: While substituting values, especially in expressions like \( -12t \), students may make calculation mistakes leading to incorrect velocity values.

Exam Tip: Always double-check your differentiation steps and ensure you substitute the correct time into the velocity expression. It's also helpful to verify units and reasonableness of the answer (e.g., negative velocity might not make sense in some contexts, but here it's acceptable).