The third graph
Plot the acceleration \(a\) against time and you get the acceleration-time graph, or the a-t graph. For the motions of this chapter it is usually the simplest of the three graphs. A body with constant acceleration has a flat a-t graph. A body that speeds up, runs steadily and then brakes has an a-t graph made of three flat steps: positive, zero and negative.
The area under the v-t graph was the displacement, because displacement is velocity added up over time. The same argument works one level down. Acceleration is the rate of change of velocity. In a thin strip of time \(dt\) the velocity changes by \(a\,dt\), which is the area of that strip under the a-t graph. Adding the strips gives the total change in velocity.
\[ \Delta v = v_2 - v_1 = \int_{t_1}^{t_2} a\,dt = \text{area under the } a\text{-}t \text{ graph} \]
Decide the answer to the question below before you press play on the figure. The area above the axis and the area below it both matter.
A car moving at \(2\,\text{m/s}\) has an acceleration of \(+2\,\text{m/s}^2\) for \(3\,\text{s}\), zero for \(2\,\text{s}\), then \(-2\,\text{m/s}^2\) for \(4\,\text{s}\). What is its velocity at the end, at \(t = 9\,\text{s}\)?
The figure shows why the area gives only the change. The area under the a-t graph starts at zero, yet the velocity starts at \(2\,\text{m/s}\). The velocity at any instant is the starting velocity plus the area so far. Without the starting velocity, the a-t graph cannot tell whether the car moved at \(2\,\text{m/s}\) or at \(20\,\text{m/s}\).
The area under an a-t graph is never the velocity itself. It is the change in velocity. Always ask for the initial velocity \(u\), and write \(v = u + \text{area}\).
Use the a-t graph of the lift. The lift starts from rest. Find its velocity at \(t = 2\,\text{s}\), \(8\,\text{s}\) and \(10\,\text{s}\), and the height it rises in the \(10\,\text{s}\).
The lift starts from rest, so \(u = 0\). Take upward as positive. The a-t graph has three steps: \(+1\,\text{m/s}^2\) from \(0\) to \(2\,\text{s}\), zero from \(2\) to \(8\,\text{s}\), and \(-1\,\text{m/s}^2\) from \(8\) to \(10\,\text{s}\).
Velocity at any time = \(0\) + area under the a-t graph up to that time.
At \(2\,\text{s}\): \(v = 1 \times 2 = 2\,\text{m/s}\). From \(2\) to \(8\,\text{s}\) the area is zero, so \(v = 2\,\text{m/s}\) at \(8\,\text{s}\). At \(10\,\text{s}\): \(v = 2 + (-1)(2) = 0\).
The v-t graph is a trapezium: up to \(2\,\text{m/s}\), flat, down to \(0\). Its area is \(\tfrac12 \times 2 \times 2 + 6 \times 2 + \tfrac12 \times 2 \times 2 = 16\,\text{m}\). The lift rises about five floors.
A lift starts from rest with an acceleration of \(1.5\,\text{m/s}^2\) for \(2\,\text{s}\). Its acceleration is then zero. What is its velocity \(5\,\text{s}\) after it started, in m/s?
The ladder of three graphs
The three graphs are linked by two operations. Going down the ladder, take slopes. The slope of the x-t graph is the velocity, and the slope of the v-t graph is the acceleration. Going up the ladder, take areas. The area under the a-t graph is the change in velocity, and the area under the v-t graph is the change in position.
The two directions are not equally complete. A slope needs nothing else. An area gives only a change, so every step up the ladder needs one starting value. To climb from the a-t graph to the x-t graph you need the initial velocity and the initial position. This is the graph form of a fact from calculus: a derivative is unique, but an integral carries a constant.
The slope of a position-time graph is the velocity, and the slope of a velocity-time graph is the acceleration. The area under an acceleration-time graph is the change in velocity, and the area under a velocity-time graph is the displacement. Areas give changes, so each step up needs a starting value.
What does the area under an acceleration-time graph between two instants give?
You are given the a-t graph of a body, its initial velocity and its initial position. Put the steps in order to reach its position at a later time.
Two cars have exactly the same a-t graph. Explain why their v-t graphs may still be different.
The a-t graph gives only the change in velocity over each interval, as an area. Two cars with the same a-t graph have the same changes, but they can start with different velocities. Their v-t graphs then have the same shape, shifted up or down by the difference in their initial velocities.
One more step down the ladder exists. The slope of the a-t graph is called the jerk, measured in m/s³. A sudden step in the a-t graph means a very large jerk, and passengers feel it as a jolt. Lift and Metro engineers round off the corners of the a-t steps so that the jerk stays small and the ride feels smooth.
A mobile phone carries an accelerometer, a tiny chip that records the acceleration many times a second. The phone therefore holds an a-t graph of your motion. Adding up areas once gives the velocity, and a second time gives the displacement. Each step up the ladder also adds up the small errors of the chip, so the estimate drifts within seconds. That is why maps still depend on GPS, which measures the position directly.
- Taking the area under an a-t graph as the velocity.The area is the change in velocity. Add the initial velocity: \(v = u + \text{area}\).
- Reading a flat a-t graph as a body at rest.A flat a-t line means constant acceleration. The body is at rest only if the velocity is zero, which the a-t graph cannot show.
- Forgetting that negative a-t area reduces the velocity.Area below the time axis counts as a negative change in velocity, exactly as on the v-t graph.