Class JEE Mathematics Practice Q #707
KNOWLEDGE BASED
APPLY
4 Marks 2023 MCQ SINGLE
A particle moves along a straight line such that its displacement x changes with time t as x = t^3 - 6t^2 + 3t + 4. What is the velocity when acceleration is zero?
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Correct Answer: -9 m/s
Explanation
Velocity v = dx/dt = 3t^2 - 12t + 3. Acceleration a = dv/dt = 6t - 12. For a=0, 6t=12 => t=2s. At t=2, v = 3(4) - 12(2) + 3 = -9.

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Step-by-Step Solution

1. Find the velocity function by differentiating the displacement function with respect to time:

x(t) = t^3 - 6t^2 + 3t + 4

v(t) = dx/dt = 3t^2 - 12t + 3

2. Find the acceleration function by differentiating the velocity function with respect to time:

a(t) = dv/dt = 6t - 12

3. Set the acceleration to zero and solve for time t:

6t - 12 = 0

6t = 12

t = 2

4. Substitute the value of t back into the velocity function to find the velocity when acceleration is zero:

v(2) = 3(2)^2 - 12(2) + 3

v(2) = 3(4) - 24 + 3

v(2) = 12 - 24 + 3

v(2) = -9

Correct Answer: -9

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the concepts of differentiation to find velocity and acceleration and then solve for velocity when acceleration is zero.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps involving differentiation and algebraic manipulation to arrive at the solution. It's about knowing 'how' to solve the problem.
Syllabus Audit: In the context of JEE, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of kinematic equations and calculus, which are core concepts in the syllabus.