A Delhi Metro train leaves a station, gains speed, runs steadily for a while and then brakes to a stop at the next station. The driver's cabin records only the speed at every second. Nobody writes down the distance. Yet that record is enough to tell exactly how far apart the two stations are.
Height and slope
Plot time along the horizontal axis and the velocity \(v\) along the vertical axis. The result is the velocity-time graph, or the v-t graph. The height of the curve at an instant is the velocity at that instant. A point above the time axis means motion in the positive direction; a point below it means motion in the negative direction.
The slope of the v-t graph is the rate of change of velocity, \(dv/dt\). That is the acceleration. A rising line means a positive acceleration and a falling line means a negative one. A flat line means zero acceleration: the velocity is constant, but it need not be zero.
A flat line means something different on the two graphs. A flat x-t graph means the body is at rest. A flat v-t graph means the body moves at a steady velocity. Before you read any graph, read the label on its vertical axis.
Part of the v-t graph of a bus is a horizontal line at \(v = 12\,\text{m/s}\). What is the bus doing during that part?
Area is displacement
For a body moving at a constant velocity \(v\) for a time \(t\), the displacement is \(v \times t\). On the v-t graph that product is the area of a rectangle of height \(v\) and width \(t\). The idea survives when the velocity changes. Cut the time into very thin strips. In each strip the velocity hardly changes, so the displacement in the strip is its height times its width. Adding the strips adds their areas.
In the language of calculus, adding infinitely many thin strips is integration, and the result is an integral. The displacement from \(t_1\) to \(t_2\) is the integral of the velocity:
\[ \Delta x = \int_{t_1}^{t_2} v\,dt = \text{area under the } v\text{-}t \text{ graph from } t_1 \text{ to } t_2 \]
The graph of the Metro run is a trapezium. Its three parts are easy to measure. The first triangle has area \(\tfrac12 \times 20 \times 20 = 200\,\text{m}\). The rectangle has area \(40 \times 20 = 800\,\text{m}\). The last triangle has area \(200\,\text{m}\) again. The total is \(1200\,\text{m}\), so the two stations are \(1.2\,\text{km}\) apart. Notice the units: a height in m/s times a width in s gives an area in metres.
The slopes of the same graph give the accelerations. The first line rises by \(20\,\text{m/s}\) in \(20\,\text{s}\), so the acceleration is \(+1\,\text{m/s}^2\). The middle line is flat, so the acceleration is zero. The last line falls by \(20\,\text{m/s}\) in \(20\,\text{s}\), so the acceleration is \(-1\,\text{m/s}^2\). One graph, read two ways, gives both the braking and the distance.
A car starts from rest and its velocity rises steadily to \(12\,\text{m/s}\) in \(4\,\text{s}\). It then runs at \(12\,\text{m/s}\) for \(6\,\text{s}\). Find the displacement in these \(10\,\text{s}\), in metres.
Area below the axis
When the v-t graph dips below the time axis, the velocity is negative. The body moves backward. The displacement in a thin strip is then a negative velocity times a positive time, which is negative. So any area that lies below the time axis counts as negative displacement. The displacement over an interval is the area above the axis minus the area below it.
The distance travelled is a different question. Distance never decreases, whichever way the body moves. So for the distance, every area counts as positive: add the area above the axis and the area below it, both taken as positive numbers.
Use the probe. For the first \(3\,\text{s}\) the ball moves up the slope and the area grows to \(\tfrac12 \times 3 \times 6 = 9\,\text{m}\). At \(t = 3\,\text{s}\) the graph crosses the axis: the ball stops for an instant at its highest point. After that, the area is below the axis and the displacement falls. At \(t = 6\,\text{s}\) the two triangles are equal, so the displacement is zero and the ball is back where it started. The distance, however, is \(9 + 9 = 18\,\text{m}\).
The velocity of a particle is \(v = 8 - 2t\) m/s. Find its displacement and the distance it travels from \(t = 0\) to \(t = 6\,\text{s}\).
The v-t graph is a straight line from \(+8\,\text{m/s}\) at \(t = 0\) to \(-4\,\text{m/s}\) at \(t = 6\,\text{s}\). It crosses the axis where \(8 - 2t = 0\), at \(t = 4\,\text{s}\).
Area above the axis, from \(0\) to \(4\,\text{s}\): a triangle, \(\tfrac12 \times 4 \times 8\). Area below the axis, from \(4\) to \(6\,\text{s}\): a triangle, \(\tfrac12 \times 2 \times 4\).
The upper triangle is \(16\,\text{m}\) and the lower triangle is \(4\,\text{m}\). The displacement is \(16 - 4 = 12\,\text{m}\). The distance is \(16 + 4 = 20\,\text{m}\).
The particle moves \(16\,\text{m}\) forward, turns at \(t = 4\,\text{s}\), and comes \(4\,\text{m}\) back. It ends \(12\,\text{m}\) from the start after travelling \(20\,\text{m}\).
Use the probe for the ball on the slope. Find the displacement of the ball from \(t = 0\) to \(t = 5\,\text{s}\), in metres.
The v-t graph of a toy car starts at \(+4\,\text{m/s}\), falls in a straight line, and reaches \(-4\,\text{m/s}\) at \(t = 8\,\text{s}\). Which statements are true?
Two Metro stations are \(1.2\,\text{km}\) apart. The train speeds up and brakes at about \(1\,\text{m/s}^2\), and its top speed is about \(22\,\text{m/s}\), that is \(80\,\text{km/h}\). Estimate the time between the two stations, in seconds.
In Paris, around 1350, the scholar Nicole Oresme drew the speed of a moving body as a row of vertical lines. The lines stood on a line of time. The tops of the lines made a figure. Oresme saw that the area of the figure measured the distance covered. For a body speeding up uniformly, his figure was a triangle. He showed that its area equals that of a rectangle at the average speed. This was the v-t graph and its area, drawn nearly three hundred years before Galileo used the same result.
- Reading a flat v-t graph as rest.On a v-t graph a flat line means constant velocity. Rest is a flat line at \(v = 0\), on the time axis itself.
- Adding the area below the axis when the displacement is asked.Area below the time axis is negative displacement. Subtract it for displacement; add it only for distance.
- Using the equations of motion across a turning point to find distance.Find where the graph crosses the axis, split the motion there, and add the areas as positive numbers.