Displacement

Question

A particle moves 30 m north, then 40 m east, and finally 50 m south. What is the magnitude of its displacement (approximately)?

RankGuru · Accepted Answer

1. **Vector Components**: Let East be $+X$ ($\hat{i}$) and North be $+Y$ ($\hat{j}$).
   - $\vec{s}_1 = 30 \hat{j}$ (North).
   - $\vec{s}_2 = 40 \hat{i}$ (East).
   - $\vec{s}_3 = -50 \hat{j}$ (South).
2. **Summing Vectors**:
   - $\vec{s}_{net} = 40 \hat{i} + (30 - 50) \hat{j} = 40 \hat{i} - 20 \hat{j}$.
3. **Calculate Magnitude**:
   - $|s| = \sqrt{40^2 + (-20)^2} = \sqrt{1600 + 400} = \sqrt{2000}$.
4. **Final Value**:
   - $|s| = 20\sqrt{5} \approx 20 	imes 2.236 \approx 44.72$ m.
5. **Approximate Answer**: **45 m**.

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