Equations Of Motion
Easyphysics
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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Solution
Incorrect! 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)ā¹a=5 m/s2.
- Calculate Distance (s): Using s=ut+21āat2
- s=0(5)+21ā(5)(5)2
- s=2.5Ć25=62.5 m.
- Alternative Method: Using average velocity for constant acceleration:
- s=(2u+vā)t=(20+25ā)Ć5=12.5Ć5=62.5 m.
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- equations of motion
- Difficulty
- Easy
- Year
- 2025
This easy difficulty physics question is from the chapter kinematics, covering the topic of equations of motion. It appeared in the 2025 exam. Practice this and similar questions to strengthen your understanding of kinematics concepts.
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