Uniformly Accelerated Motion
A train starting from rest accelerates uniformly and covers 200 m in the first 10 seconds along a straight track, what is the acceleration of the train?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- uniformly accelerated motion
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
4m/s2
As described in NCERT Class 11, Chapter 3 (Motion in a Straight Line), for uniformly accelerated motion starting from rest the displacement is given by s = ut + (1/2)at², where u is the initial velocity. Since the train starts from rest, u = 0, so the equation reduces to s = (1/2)at². Substituting s = 200 m and t = 10 s gives 200 = (1/2) × a × 100 = 50a. Solving, a = 200/50 = 4 m/s². The option 2 m/s² is wrong because it omits the factor of one-half and incorrectly uses s = at²/4. The option 20 m/s² results from treating distance as if equal to velocity times time without the kinematic factor. The option 1 m/s² underestimates by misapplying the formula. Plausibility check: with a = 4 m/s², the final velocity is at = 40 m/s and the average velocity is 20 m/s, which over 10 s gives exactly 200 m, confirming consistency.
This easy difficulty physics question is from the chapter kinematics, covering the topic of uniformly accelerated motion. It appeared in the 2025 exam.
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