Instantaneous Acceleration From Calculus
The position of a particle along a straight line varies with time as x equals 2t cubed minus 3t squared plus 4, what is its acceleration at t equals 2 seconds?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- instantaneous acceleration from calculus
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
18m/s2
As developed in NCERT Class 11, Chapter 3 (Motion in a Straight Line), instantaneous velocity is the first derivative of position with respect to time, and instantaneous acceleration is the second derivative. Given x = 2t³ − 3t² + 4, differentiating once gives velocity v = dx/dt = 6t² − 6t. Differentiating again gives acceleration a = dv/dt = 12t − 6. Substituting t = 2 s yields a = 12 × 2 − 6 = 24 − 6 = 18 m/s². The option 12 m/s² wrongly uses only the leading term 12t without the constant. The option 24 m/s² stops at 12t = 24 and ignores the −6. The option 6 m/s² results from differentiating only once too few times or arithmetic error. Plausibility check: the velocity at t = 2 is 6(4) − 6(2) = 12 m/s, and the acceleration steadily increases with time as 12t − 6, so at t = 2 a positive 18 m/s² is dimensionally and physically consistent with the cubic position function.
This hard difficulty physics question is from the chapter kinematics, covering the topic of instantaneous acceleration from calculus. It appeared in the 2025 exam.
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