Position-Time Graphs — Practice Problem Solutions

1. Slope $=\dfrac{12-0}{4-0}=3\text{ m/s}$ — the velocity is $3\text{ m/s}$.

2. A straight line starting at the origin $(0,0)$, rising with a constant positive slope of 3, ending at the point $(5\text{ s},15\text{ m})$.

3. Slope from $t=0$ to $t=4\text{ s}$: $\dfrac{8-0}{4-0}=2\text{ m/s}$.

4. From $t=0$ to $t=2\text{ s}$: $\dfrac{6-0}{2-0}=3\text{ m/s}$. From $t=2\text{ s}$ to $t=4\text{ s}$: $\dfrac{6-6}{4-2}=0\text{ m/s}$. The object moved at a steady 3 m/s for the first two seconds, then came to a complete stop and stayed there for the next two.

5. A flat line at the starting position for the first 2 s, then a straight line with constant negative slope for the next 3 s, ending at a lower (more negative) position than where it started.

6. a) 0–2 s: $\dfrac{10-0}{2}=5\text{ m/s}$. 2–4 s: $\dfrac{10-10}{2}=0\text{ m/s}$. 4–6 s: $\dfrac{4-10}{2}=-3\text{ m/s}$. b) The 0–2 s interval, since it has the largest slope magnitude (5 m/s, versus 0 and 3 m/s) — not because the line "looks" steepest, but because its calculated slope value is the biggest. c) The object moved forward 10 m in the first 2 s, stayed completely still for the next 2 s, then moved backward 6 m (from the 10 m mark to the 4 m mark) over the final 2 s.

7. The same total displacement ($4\text{ m/s}\times10\text{ s}=40\text{ m}$) covered in half the time means a steeper slope: $\dfrac{40\text{ m}}{5\text{ s}}=8\text{ m/s}$ — twice as steep as the original.