Motion in a Straight Line · Lesson 9 of 10
Interpreting Complex Motion Graphs
“Build and test piecewise graphs, recognise impossible graphs and analyse rebounds and oscillatory motion.”
• Construct position-time graphs for journeys with stops and reversals. • Distinguish valid graphs from impossible descriptions of one particle. • Interpret slope, curvature and equal-time intervals. • Build speed-time and velocity-time descriptions of a bouncing ball.
A graph-reading routine
First read both axis labels and units. Then decide what height, slope and area mean for those particular axes. On x–t, height is position and slope is velocity. On v–t, height is velocity, slope is acceleration and area is displacement. On speed–time, area is distance; its slope is the rate of change of speed, which equals signed acceleration only when positive velocity is maintained. Do not transfer an interpretation from one graph type to another.
A journey with a long stop
Problem
A person leaves at 9:00, walks 2.5 km at 5 km/h, stays at the office until 17:00, then returns at 25 km/h. Construct the position-time graph.
- 1.Let t be hours after 9:00, x be kilometres from home toward the office. The outward journey ends after 2.5/5 = 0.5 h, at 9:30.
- 2.The office stay ends at t = 8 h. The return takes 2.5/25 = 0.1 h = 6 min, ending at 17:06.
- 3.Plot (0,0), (0.5,2.5), (8,2.5) and (8.1,0), joining consecutive points by straight lines.
- 4.The slopes are +5 km/h, 0 and −25 km/h. The long horizontal segment is rest, not a long spatial route. Use an enlarged inset or annotate the brief return because a full-day horizontal scale compresses it.
Repeated forward and backward steps
Repeated cycles can be simplified, but only after checking whether the destination is reached during a cycle. Net progress per cycle is not the same as instantaneous progress. A walker may reach a target during the forward part before the backward part would begin. Stop the calculation at the first arrival, rather than forcing an extra complete cycle.
Problem
A walker takes five 1 m forward steps, then three 1 m backward steps, repeating this pattern. Each step takes 1 s. When is a point 13 m ahead first reached?
- 1.Each complete 8 s cycle gives net progress 5 − 3 = 2 m.
- 2.After four cycles, t = 32 s and x = 8 m. The forward peaks in those cycles were 5, 7, 9 and 11 m, so the target has not been reached.
- 3.The next five forward steps reach 13 m at t = 37 s. There is no subsequent backward segment because the question ends at first arrival.
- 4.An x–t graph has alternating slopes +1 and −1 m/s. Using average net progress 2/8 m/s all the way to 13 m would predict 52 s and miss the first arrival.
Reading two travellers on the same graph
A larger final x-coordinate indicates a more distant home when the school is x = 0 and both homes lie in the positive direction. An earlier departure is shown by an earlier start time, not by a steeper slope. A steeper rising line means a faster walk. Intersections indicate equal positions at the same time, and a crossing of the lines indicates one traveller overtakes the other.
For the graph of children A and B, A’s home P is closer than B’s home Q. A leaves earlier; B’s steeper line shows faster walking. B crosses A once and reaches the more distant home earlier, so their arrival times differ. These conclusions require comparing horizontal coordinates for times and vertical coordinates for positions, not just where a line ends visually.
Graphs that cannot describe one particle
| Graph feature | Why it is impossible or invalid |
|---|---|
| A loop in an x–t curve | At some times it assigns two positions to one particle. |
| A loop in a v–t curve | At some times it assigns two velocities to the same particle. |
| A speed curve below zero | Speed is a magnitude and cannot be negative. |
| A total path-length curve that decreases | Accumulated distance never decreases; returning adds distance. |
A curve that is horizontal for t < 0 and parabolic for t > 0 can describe an object at rest until a chosen instant, followed by uniformly accelerated motion along the same straight line. It does not imply a straight spatial path before zero and a curved spatial path after zero. Likewise, negative t is allowed if the clock zero is a selected event rather than the beginning of observation.
Inventing physical stories from graph shapes
An x–t graph that is initially flat, rises linearly, falls through x = 0 to a negative constant value and becomes flat again can describe a person resting, walking forward, turning and walking back past the start, then resting. Changes between straight segments idealise brief periods of acceleration. A sequence of v–t straight lines with constant negative slope and upward jumps can describe a bouncing ball with upward positive. A narrow positive pulse on an a–t graph can describe a brief push: velocity rises during the pulse and is constant before and after it. The area of the pulse, not its height alone, gives the velocity change.
Signs from an oscillating position graph
For an oscillating x–t graph, read three independent features. Above or below zero gives the sign of position; rising or falling gives the sign of velocity; upward or downward concavity gives the sign of acceleration. A maximum has zero slope and negative acceleration if it is rounded downward; a minimum has zero slope and positive acceleration if it is rounded upward. The detailed theory of oscillations is not needed to read these local features.
| Time in the oscillation plot | Position | Velocity (slope) | Acceleration (curvature) |
|---|---|---|---|
| 0.3 s | Negative | Negative | Positive |
| 1.2 s | Positive | Positive | Negative |
| −1.2 s | Negative | Positive | Positive |
These signs come from the plotted curve. For example, at 0.3 s the particle lies below zero and is moving farther in the negative direction, but the curve bends upward. Thus positive acceleration is reducing the magnitude of its negative velocity. Position alone would not reveal its direction of travel.
Comparing equal time windows
On an x–t graph, average velocity is endpoint change divided by window width. Average speed uses total vertical travel along the position coordinate, splitting at any turning point. For equal widths, compare those changes directly. A small net displacement near a peak may conceal some out-and-back motion, so inspect the whole interval. In the three marked position windows above, interval 3 has greatest average speed and interval 2 least; average velocity is positive in 1 and 2, negative in 3.
On the speed-time graph above, motion remains in a constant positive direction. Average acceleration is therefore endpoint speed difference divided by window width; interval 2 has the greatest magnitude. Average speed comes from area divided by width; interval 3 has the largest average speed. Velocity is positive in all three intervals; acceleration is positive in interval 1 and negative in intervals 2 and 3. At the smooth extrema A, B, C and D, the tangent is horizontal and instantaneous acceleration is zero. A large speed does not imply a large acceleration.
Building a bouncing-ball graph
Between floor contacts, a ball undergoes free fall. At contact, the floor exerts a force for a short time and reverses its velocity. An ideal graph treats that short event as a jump. If each rebound retains 90% of the speed just before impact, upward rebound speed is 0.9 times the downward arrival speed. It is speed, not velocity, that loses 10%: signed velocity also changes direction.
Problem
Drop a ball from 90 m and retain 90% of impact speed at each bounce. Use g = 9.8 m/s² and ignore air resistance.
- 1.First impact occurs at t₁ = √(2 × 90/9.8) = 4.286 s. Arrival speed is g t₁ = 42.0 m/s.
- 2.Just after the first bounce the speed is 0.9 × 42 = 37.8 m/s upward. It takes 37.8/9.8 = 3.857 s to reach the top, at absolute time 8.143 s.
- 3.The return from the top takes the same time, so the next impact is at 4.286 + 2 × 3.857 = 12.000 s. The pre-impact speed is 37.8 m/s and post-impact speed is 34.02 m/s.
- 4.Plot speed rising linearly from (0,0) to (4.286,42), dropping abruptly to 37.8, falling linearly to (8.143,0), then rising to (12,37.8). At exactly 12 s distinguish the pre-impact and post-impact values.
- 5.With upward positive, signed velocity before the first impact is −42 m/s and after it is +37.8 m/s. Between impacts v has slope −g, whereas the speed graph has slope −g on ascent and +g on descent.
A rebound with retained speed fraction e reaches a height e² times the preceding drop height, since H = u²/(2g). Here the first rebound rises 0.9² × 90 = 72.9 m above the floor. Its round-trip flight time is 2u/g. At the top, speed is zero but gravitational acceleration is not. At idealised collision instants, there is no single finite acceleration compatible with the velocity jump; a real contact has a large acceleration over a brief duration.
Quiz
Which graph feature is impossible for total distance travelled versus time?
Key Takeaways
• Read axis labels before interpreting height, slope or area. • A time graph is a record of motion, not the spatial path. • One particle has one position and one velocity at each ordinary instant. • Distance is non-decreasing, and speed is non-negative. • At a bounce, signed velocity reverses; speed may decrease by a specified fraction.