Coaching Centre
Analysing motion graphs
Moving between position-time, velocity-time and acceleration-time graphs using gradients and areas, distance versus displacement from a velocity-time graph, and interpreting the shapes of graphs.
A lift in a Sydney office tower speeds up, cruises, then slows to a stop at the 30th floor. You can feel two of those three stages in your stomach. Each stage has a different shape on each of three graphs, and HSC questions love asking you to draw one from another.
You met graph matching in Year 10. In Year 11 the graphs carry numbers: you read gradients and areas off them, and convert one graph into another. Two rules do almost all the work.
Two rules: gradient and area
| Graph | Gradient gives | Area under it gives |
|---|---|---|
| position–time ($s$–$t$) | velocity | (nothing useful) |
| velocity–time ($v$–$t$) | acceleration | displacement |
| acceleration–time ($a$–$t$) | (not needed in this course) | change in velocity |
Going down the table (s → v → a), take gradients. Going up (a → v → s), take areas, then add to the starting value.
Reading the shapes
- On an $s$–$t$ graph, a straight line means constant velocity. A curve getting steeper means speeding up; a curve flattening off means slowing down.
- On a $v$–$t$ graph, a straight line means constant acceleration. A horizontal line means constant velocity (zero acceleration), not at rest unless it is on the axis.
- Where a $v$–$t$ graph crosses the time axis, the object stops and reverses. On the $s$–$t$ graph that is a turning point (a peak or a trough).
Areas below the axis: displacement and distance
Area under a $v$–$t$ graph below the time axis is negative displacement: the object is moving backwards. So:
- displacement = area above the axis − area below the axis
- distance travelled = area above + area below (all counted as positive)
A remote-control car starts at 3.0 m s⁻¹ backwards ($u = -3.0$ m s⁻¹) and accelerates at a constant +1.5 m s⁻² for 6.0 s. Find its displacement and the distance it travels.
A lift starts from rest. Its acceleration is +1.0 m s⁻² for 2.0 s, zero for 6.0 s, then −1.0 m s⁻² for 2.0 s. Find its top speed and how far it travels.
- Choose Car and drag the time cursor to 10 s. The car ends at rest after a displacement of 40 m. Find the 40 m as an area on the v–t graph.
- Choose Reversing ($u = -3$ m/s). At what time does the s–t graph have its lowest point? Check that the v–t graph crosses the axis at that time.
- Compare the distance and displacement readouts for the reversing car. When do they start to differ?
- In Your own, make an a–t graph that returns the object to where it started (displacement 0) at 10 s.
Test yourself
On a velocity–time graph, the area between the line and the time axis represents
Velocity × time = displacement.
An object's position–time graph is a straight line sloping downwards. The object has
A straight line has a constant gradient, and a downward slope is negative.
A velocity–time graph is a straight line from +6 m s⁻¹ at 0 s to −6 m s⁻¹ at 4 s. The displacement over the 4 s is
The triangles above and below the axis have equal areas (6 m each), so they cancel. The distance is 12 m.
At the moment a $v$–$t$ graph crosses the time axis, the object
$v = 0$ there, and the sign of $v$ changes. The displacement is a maximum or minimum only from that reversal onwards, not necessarily the largest overall.
Label what each feature of the graphs tells you.
A ball is thrown up and caught at the same height. Put its v–t graph features in time order.
The gradient of a position–time graph is the . The area under an acceleration–time graph is the change in . A horizontal line on a velocity–time graph means zero . Area below the time axis on a velocity–time graph counts as displacement.
A cyclist starts from rest, accelerates uniformly to 8.0 m s⁻¹ in 4.0 s, rides at 8.0 m s⁻¹ for 10 s, then brakes uniformly to rest in 2.0 s. (a) Describe the shape of the velocity–time graph. (b) Calculate the acceleration in each stage. (c) Calculate the total displacement and the average velocity.
(a) A straight line rising from 0 to 8.0 m s⁻¹ over 0–4.0 s, a horizontal line at 8.0 m s⁻¹ from 4.0 to 14 s, then a straight line falling to 0 at 16 s.
(b) $a_1 = 8.0/4.0 = 2.0$ m s⁻²; $a_2 = 0$; $a_3 = (0 - 8.0)/2.0 = -4.0$ m s⁻².
(c) Areas: $\tfrac12(4.0)(8.0) = 16$ m, $(10)(8.0) = 80$ m, $\tfrac12(2.0)(8.0) = 8.0$ m. Total displacement $= 104$ m forwards. Average velocity $= 104/16 = 6.5$ m s⁻¹ forwards.