Rectilinear Kinematics
Average Velocity & Average Speed — Multi-Segment Journey
A particle travels along a straight-line path in two segments.
In the first 4 s it moves from position sA = −8 m
to position sB = +3 m.
In the next 5 s it moves from sB to sC = −6 m.
Determine the particle’s average velocity and average speed over the full 9-second interval.
What We Know
- Segment A → B: from s = −8 m to s = +3 m in 4 s
- Segment B → C: from s = +3 m to s = −6 m in 5 s
- Total elapsed time: 4 + 5 = 9 s
- Find: average velocity over the full 9 s
- Find: average speed over the full 9 s
Key distinction:
Average velocity = net displacement / total time — uses only start and end positions, signed.
Average speed = total distance traveled / total time — counts every metre walked, unsigned.
These two will differ whenever the particle changes direction, which it does here: it moves right (A→B) and then left (B→C).
Step 1
Visualise the Path on a Number Line
Plotting the three positions on the straight-line axis clarifies the motion before any calculation:
POSITION NUMBER LINE
→ 11 m (A → B, 4 s)
← 9 m (B → C, 5 s)
C
s = −6 m
s = −6 m
end
s = 0
A
s = −8 m
s = −8 m
start
B
s = +3 m
s = +3 m
turn
← Negative
Net displacement = +2 m | Total distance = 20 m
Positive →
Key observation: The particle travels right from A to B, then reverses and travels left from B to C. It ends up only 2 m to the right of where it started — a modest net displacement — but it covered a total path length of 20 m.
Part A — Average Velocity
Step 2
Compute Net Displacement and Average Velocity
Average velocity depends only on the starting and ending positions — the path in between is irrelevant:
\[ \Delta s = s_C – s_A = -6 – (-8) = +2 \ \text{m} \]
\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{+2 \ \text{m}}{4 \ \text{s} + 5 \ \text{s}} = \frac{2}{9} \]
\[ \bar{v} = 0.222 \ \text{m/s} \quad \text{(directed to the right)} \]
Physical meaning: The positive sign confirms the particle ends up to the right of where it began. However, the small magnitude (0.222 m/s) reflects the fact that most of the rightward gain from A→B was cancelled out by the leftward return from B→C — only 2 m of net progress in 9 s.
Part B — Average Speed
Step 3
Compute Distance for Each Segment
Distance is always positive — it is the actual path length walked, regardless of direction:
\[ d_{AB} = |s_B – s_A| = |3 – (-8)| = |{+11}| = 11 \ \text{m} \]
\[ d_{BC} = |s_C – s_B| = |-6 – 3| = |{-9}| = 9 \ \text{m} \]
\[ d_\text{total} = d_{AB} + d_{BC} = 11 + 9 = 20 \ \text{m} \]
Why not just |sC − sA|? Because the particle changed direction at B. Using only start and end would give |−6 − (−8)| = 2 m — which dramatically undercounts the actual path walked. Whenever direction reverses, each segment must be computed and summed separately.
Step 4
Compute Average Speed
Average speed is total distance divided by total elapsed time:
\[ \bar{v}_\text{speed} = \frac{d_\text{total}}{\Delta t} = \frac{20 \ \text{m}}{9 \ \text{s}} \]
\[ \bar{v}_\text{speed} = 2.22 \ \text{m/s} \]
| Quantity | Segment A→B | Segment B→C | Total (9 s) |
|---|---|---|---|
| Time elapsed | 4 s | 5 s | 9 s |
| Distance traveled | 11 m | 9 m | 20 m |
| Displacement | +11 m | −9 m | +2 m |
Velocity vs. Speed — the contrast:
Average velocity = +0.222 m/s (small — net motion nearly cancelled).
Average speed = 2.22 m/s (10× larger — reflects the full journey).
This factor-of-10 difference perfectly illustrates why these two quantities must never be confused in kinematics.
Final Answers
AVERAGE VELOCITY (over 9 s)
v̄ = +2/9 ≈ 0.222 m/s → (to the right)
AVERAGE SPEED (over 9 s)
v̄speed = 20/9 ≈ 2.22 m/s