1.4 Using Areas to Find Distances and Displacements
The following graphs model the motion of a stone falling from the rest.
Figure 1.14
Figure 1.15
Finding the area under speed–time graphs
Many of these graphs consist of straight-line sections. The area is easily found by splitting it up into triangles, rectangles or trapezia.
Figure 1.16
Example 1.4.1:
The graph in Figure 1.17 shows Hinesh’s journey from the time he turns on to the main road until he arrives home. How far does Hinesh cycle?
Figure 1.17
Solution 1.4.1:
Split the area under the speed–time graph into three regions.
Hinesh cycles 262m.
The area between a velocity–time graph and the time axis
Example 1.4.2:
Sunil walks east for 6 s at 2 m s−1 then west for 2 s at 1 m s−1. Draw
(i) a diagram of the journey
(ii) the speed–time graph
(iii) the velocity–time graph.
Interpret the area under each graph.
Solution 1.4.2:
Figure 1.18
Figure 1.19
Figure 1.20
The area between a velocity–time graph and the time axis represents the change in position, that is the displacement.
When the velocity is negative, the area is below the time axis and represents a displacement in the negative direction, west in this case.
Example 1.4.3:
On the London underground, Oxford Circus and Piccadilly Circus are 0.8 km apart. A train accelerates uniformly to a maximum speed when leaving Oxford Circus and maintains this speed for 90 s before decelerating uniformly to stop at Piccadilly Circus. The whole journey takes 2 minutes. Find the maximum speed.
Solution 1.4.3:
Figure 1.22
The above sketch of the speed–time graph of the journey shows the given information, with suitable units. The maximum speed is v m s−1.
The maximum speed of the train is 7.6 m s−1 (to 2 s.f.).