Displacement
A particle moves 30 m north, then 40 m east, and finally 50 m south. What is the magnitude of its displacement (approximately)?
Select the correct option:
Solution
45 m
- Vector Components: Let East be +X (i^) and North be +Y (j^).
- s1=30j^ (North).
- s2=40i^ (East).
- s3=−50j^ (South).
- Summing Vectors:
- snet=40i^+(30−50)j^=40i^−20j^.
- Calculate Magnitude:
- ∣s∣=402+(−20)2=1600+400=2000.
- Final Value:
- ∣s∣=205≈20×2.236≈44.72 m.
- Approximate Answer: 45 m.
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- displacement
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
45 m
- Vector Components: Let East be +X (i^) and North be +Y (j^).
- s1=30j^ (North).
- s2=40i^ (East).
- s3=−50j^ (South).
- Summing Vectors:
- snet=40i^+(30−50)j^=40i^−20j^.
- Calculate Magnitude:
- ∣s∣=402+(−20)2=1600+400=2000.
- Final Value:
- ∣s∣=205≈20×2.236≈44.72 m.
- Approximate Answer: 45 m.
This medium difficulty physics question is from the chapter kinematics, covering the topic of displacement. It appeared in the 2025 exam.
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