Equations Of Motion
A car accelerates uniformly from rest to a velocity of 25 m/s in 5 seconds. What is the distance covered during this time?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- equations of motion
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
62.5 m
- Identify Given Data:
- Initial velocity u=0 (from rest).
- Final velocity v=25 m/s.
- Time t=5 s.
- Find Acceleration (a): Using v=u+at
- 25=0+a(5)⟹a=5 m/s2.
- Calculate Distance (s): Using s=ut+21at2
- s=0(5)+21(5)(5)2
- s=2.5×25=62.5 m.
- Alternative Method: Using average velocity for constant acceleration:
- s=(2u+v)t=(20+25)×5=12.5×5=62.5 m.
This easy difficulty physics question is from the chapter kinematics, covering the topic of equations of motion. It appeared in the 2025 exam.
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