Displacement Physics Example 1

Solution

1. 1
   Total distance traveled is the sum of path lengths: $d = 4 + 3 = 7 \text{ m}$.
2. 2
   Displacement is the straight-line distance from start to finish: $$|\vec{d}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m}$$
3. 3
   Direction: $\theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.9°$ north of east.