Distance, Displacement, Vectors, and Scalars — Practice Problem Solutions

1. Both legs are in the same direction (east), so distance $=12+5=17\text{ m}$ and displacement $=17\text{ m east}$ — they're equal here only because the motion never reverses direction.

2. Mass — scalar. Velocity — vector. Time — scalar. Force — vector.

3. Distance $=400+250=650\text{ m}$. Displacement: $400\text{ m east}-250\text{ m east}=150\text{ m east}$.

4. Net floor change $=+3+2-4=+1$ floor (one floor higher than the start). This is a vector quantity here — "up 1 floor" carries direction (up vs. down) along the elevator's one axis of motion, not just a magnitude.

5. The 3 km north and 3 km south cancel completely. Distance $=3+2+3=8\text{ km}$. Displacement $=2\text{ km east}$ (only the east-west leg survives).

6. Example: walk 100 m east, then 100 m west back to the start. Distance $=100+100=200\text{ m}$; displacement $=0\text{ m}$. The claim is false — displacement really is always zero when you return to your exact starting point, but distance traveled is not; it just adds up whatever path length you covered, win or lose.

7. Distance $=5+3+2=10\text{ m}$. Displacement $=5+3-2=6\text{ m east}$. The robot's direction changed exactly once — after moving 8 m east total, it reversed and moved 2 m west.

8. (B) — recording GPS coordinates at the start and end and calculating the straight-line distance between them measures displacement directly. (A), (C), and (D) all measure or relate to distance traveled (or nothing useful), not net position change.