Throw a tennis ball straight up in the corridor of your school and catch it. The ball goes up one straight line and comes down the same line. Yet if you plot its height against time, you get a smooth arch. If you let it bounce on the floor instead, the graph grows sharp corners. Real motions leave their own marks on the graphs.
A ball thrown straight up
Take upward as positive and \(g = 10\,\text{m/s}^2\). A ball leaves the hand at \(20\,\text{m/s}\). Its height is \(y = 20t - 5t^2\), its velocity is \(v = 20 - 10t\), and its acceleration is \(a = -10\,\text{m/s}^2\) at every instant. The ball reaches the top, \(20\,\text{m}\), at \(t = 2\,\text{s}\), and returns to the hand at \(t = 4\,\text{s}\).
The height-time graph is a dome, symmetric about \(t = 2\,\text{s}\). It is concave down throughout, because the acceleration is negative throughout. The v-t graph is a single straight line of slope \(-10\,\text{m/s}^2\). It crosses the time axis at \(2\,\text{s}\), exactly below the top of the dome. The a-t graph is a flat line at \(-10\,\text{m/s}^2\), with no break at the top.
The v-t graph does not bend or break at the top. That is the graph telling you that the top is not special for the acceleration. Gravity pulls the ball down at the same rate on the way up, at the top and on the way down. The areas under the v-t line check the symmetry. The triangle above the axis, \(\tfrac12 \times 2 \times 20 = 20\,\text{m}\), is the rise. The triangle below, also \(20\,\text{m}\), is the fall.
Which is the a-t graph of a ball thrown straight up, from launch until it is caught, with upward positive?
A bouncing ball
Drop a ball from \(5\,\text{m}\). It falls for \(1\,\text{s}\) and hits the floor at \(10\,\text{m/s}\). A real ball loses some speed in the bounce, so suppose it leaves the floor at \(8\,\text{m/s}\). It rises to \(8^2/20 = 3.2\,\text{m}\), falls back, and hits the floor again at \(2.6\,\text{s}\). This time it leaves at \(6.4\,\text{m/s}\). Between bounces the ball is in free fall, so each flight has the graphs you have just seen.
Look at the v-t graph first. During every flight the line falls with the same slope, \(-10\,\text{m/s}^2\), so all the flight lines are parallel. At each bounce the velocity changes from \(-10\) to \(+8\,\text{m/s}\) in a very short time. On the scale of the graph that change looks like a vertical jump. The ball is in contact with the floor for about a hundredth of a second.
Now the height-time graph. Each flight is a dome, and the domes meet the floor in sharp corners. A corner on an x-t graph means the slope, the velocity, changes suddenly. That is the jump you saw on the v-t graph. A smooth throw gives a smooth dome; a bounce gives a corner.
The a-t graph is a flat line at \(-10\,\text{m/s}^2\) during every flight, and a tall, narrow spike upward at every bounce. The area of each spike is the change in velocity, \(+18\,\text{m/s}\) at the first bounce. So the spike must be very tall, because it is very narrow.
The ball hits the floor at \(10\,\text{m/s}\) and leaves at \(8\,\text{m/s}\). The contact lasts about \(0.01\,\text{s}\). Estimate the average acceleration of the ball during the contact, in m/s².
A ball leaves the floor after a bounce at \(6\,\text{m/s}\), straight up. How high does it rise? Take \(g = 10\,\text{m/s}^2\) and give the answer in metres.
A car that stops and returns
A car moving at \(10\,\text{m/s}\) brakes and stops in \(2\,\text{s}\). It waits for \(1\,\text{s}\). Then it reverses: it speeds up backward for \(2\,\text{s}\) to \(5\,\text{m/s}\) and brakes again for \(2\,\text{s}\), stopping exactly where it started. The motion has four stages, and each stage has a constant acceleration.
Read the stages across the three graphs.
- Braking forward, \(0\) to \(2\,\text{s}\). The v-t line falls from \(10\) to \(0\). The a-t graph is \(-5\,\text{m/s}^2\). The x-t graph rises and flattens, reaching \(10\,\text{m}\).
- Waiting, \(2\) to \(3\,\text{s}\). The v-t graph lies on the axis and the a-t graph is zero. The x-t graph is flat at \(10\,\text{m}\).
- Reversing, speeding up, \(3\) to \(5\,\text{s}\). The v-t line goes below the axis, to \(-5\,\text{m/s}\). The a-t graph is \(-2.5\,\text{m/s}^2\). The x-t graph falls more and more steeply.
- Braking backward, \(5\) to \(7\,\text{s}\). The v-t line rises to zero. The a-t graph is \(+2.5\,\text{m/s}^2\). The x-t graph flattens and reaches \(0\) at \(7\,\text{s}\).
Two stages have a negative acceleration, but only one of them is a braking. In the third stage the velocity and the acceleration are both negative, so the car speeds up backward. In the last stage the acceleration is positive, yet the car slows down, because it is moving backward. The a-t graph alone never tells you whether the body speeds up; you need the v-t graph beside it.
The v-t graph has a triangle of area \(10\,\text{m}\) above the axis and a triangle of area \(10\,\text{m}\) below it. The displacement is \(10 - 10 = 0\), and the car is back at the start. The distance is \(10 + 10 = 20\,\text{m}\).
In the stop-and-return, the reversing car reached \(5\,\text{m/s}\) backward. Suppose it had reached \(6\,\text{m/s}\) instead, in the same \(2\,\text{s}\), and braked to rest in another \(2\,\text{s}\). How far from the start would it stop, in metres? Give a positive number.
For the stop-and-return car, which statements are true?
A real car cannot change its acceleration in an instant either, so its a-t graph has sloping sides instead of vertical steps. The steps of this lesson are an idealisation, just as the instant bounce was. The idealisation changes the graphs only near the joins, and the areas, which give the velocities and the positions, hardly change at all.
- Drawing the a-t graph of a thrown ball with a break at the top.Gravity acts throughout the flight. The a-t graph is one flat line at \(-g\).
- Drawing the v-t graph of a bounce as a V that touches zero.The velocity jumps from negative to positive at the bounce; the flights are parallel lines of slope \(-g\).
- Deciding speeding up or slowing down from the a-t graph alone.Read the v-t graph too. Same signs of \(v\) and \(a\) mean speeding up; opposite signs mean slowing down.