Review (Topics 1.1–1.3 Self-Check) — Practice Problem Solutions

1. Distance vs. Displacement: Distance $=5\text{ km}+2\text{ km}=7\text{ km}$. Displacement: the 2 km west partially cancels the 5 km east, leaving $5-2=3\text{ km east}$.

2. Vectors vs. Scalars: (a) "12 m/s" — scalar (a speed, no direction given). (b) "12 m/s north" — vector (a velocity). (c) "30 minutes" — scalar (a duration). (d) "3 km southeast" — vector (a displacement).

3. Average Velocity and Average Speed: Total distance $=20+20=40\text{ km}$, total time $=2\text{ h}$, so average speed $=20\text{ km/h}$. Total displacement $=0$ (the cyclist ends back where they started), so average velocity $=0$. They differ because average speed only cares about the ground covered, while average velocity only cares about net position change — a round trip covers plenty of ground but ends with zero net displacement.

4. Acceleration: $a=\dfrac{2-6}{4}=-1\text{ m/s}^2$. It's negative because velocity is decreasing. Since the skater is moving in the positive direction and slowing down, a negative acceleration here means slowing down (decelerating), not speeding up in reverse.

5. Position-Time Graph: The graph rises with a constant positive slope for the first 10 s (walking away at a steady pace), is flat for the next 5 s (standing still talking), then falls back toward the starting position with a slope twice as steep (in magnitude) as the first segment, since the return trip covers the same distance in half the time.

6. Velocity-Time Graph: (a) $a=\dfrac{12-0}{6}=2\text{ m/s}^2$. (b) Displacement during the last 4 s $=v\times t=12\times4=48\text{ m}$.