A Delhi Metro train records only its speed, once every second, in a data logger. An engineer at the depot plots the record as a velocity-time graph. From that one line she reads how hard the train accelerated and how fast it ran. She also reads how hard it braked and how far apart the stations are.
The height and the slope
The first reading of a graph is its height. On a v-t graph the height at a time \(t\) is the velocity at that instant. The second reading is the slope, the rise per unit run. The slope of any graph of \(y\) against \(x\) is the rate \(dy/dx\) at which \(y\) changes as \(x\) grows. On a straight piece, the slope is the rise divided by the run. On a curve, it is the slope of the tangent, which the chapter Differentiation explains.
The area under a graph
The third reading is the area between the graph and the horizontal axis. Cut the area into thin strips of width \(dx\). Each strip has an area \(y\,dx\), a value of \(y\) times a small step of \(x\). Adding the strips gives the running total \(\int y\,dx\). On a v-t graph each strip is \(v\,dt\), the displacement in the time \(dt\). So the area under a v-t graph is the displacement.
On a graph of \(y\) against \(x\), the height is \(y\) and the slope is the rate \(dy/dx\). The area between the graph and the \(x\) axis is the total \(\int y\,dx\), with area below the axis counted as negative. The intercepts give starting values and thresholds, and the turning points, where the slope is zero, give the greatest and least values.
The three readings change in different ways as the cursor moves. The height climbs, stays at \(20\,\text{m/s}\) and falls. The slope jumps from one value to the next at the two corners. The area grows all the time, fastest where the graph is highest. The distance between the stations is the whole area: \(200 + 800 + 200 = 1200\,\text{m}\).
The Metro train of the figure runs at \(20\,\text{m/s}\) from \(t = 20\,\text{s}\) to \(t = 60\,\text{s}\). How far does it travel while it cruises, in metres?
The units say what slope and area mean
Before reading a graph, ask two questions about its units. What is the unit of \(y\) divided by the unit of \(x\)? That is the unit of the slope. What is the unit of \(y\) times the unit of \(x\)? That is the unit of the area. If either product or quotient is a familiar physical quantity, that is what the slope or the area measures.
For which of these graphs does the area under the graph have a standard physical meaning?
Intercepts, asymptotes and turning points
An intercept is a point where the graph meets an axis. The intercept on the vertical axis is often a starting value, such as the initial velocity \(u\) on a v-t graph. The intercept on the horizontal axis is often a moment or a threshold. Examples are the time a body stops and the threshold frequency of a metal. A v-t graph that crosses the time axis marks the instant the body turns back.
An asymptote is a value that a graph approaches but never reaches. A steel ball dropped into oil speeds up less and less, and its v-t graph flattens towards the terminal velocity. A turning point is a point where the slope is zero, so the graph is momentarily flat. It may be a greatest value, such as the top of a thrown ball's flight. It may also be a least value, such as the bottom of a potential energy well.
The graph of the potential energy \(U(x)\) of a particle has a turning point at \(x = 3\,\text{m}\), where it is lowest. The force on the particle is \(F = -dU/dx\). What is the force at \(x = 3\,\text{m}\)?
The v-t graph of a steel ball falling in oil rises and flattens towards a horizontal asymptote at \(v_T\). What does the flattening say about the ball?
The work under a p-V path
The same reading works on graphs far from motion. A gas that expands against a piston does the work \(\int p\,dV\). On a graph of \(p\) against \(V\), each strip \(p\,dV\) is a thin area. So the work done by the gas is the area under its path. With the pressure in kPa and the volume in litres, one kPa L is one joule.
A gas expands from \(1\,\text{L}\) to \(5\,\text{L}\) while its pressure falls linearly from \(300\,\text{kPa}\) to \(100\,\text{kPa}\). How much work does it do, in joules?
Back to physics: a graph that crosses the axis
When a graph crosses the horizontal axis, the area below the axis counts as negative. On a v-t graph such area is displacement in the negative direction. So the displacement is the signed total of the areas. The distance travelled is the total of their sizes, all counted positive.
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 with intercept \(8\,\text{m/s}\) and slope \(-2\,\text{m/s}^2\). It runs from \(+8\,\text{m/s}\) at \(t = 0\) to \(-4\,\text{m/s}\) at \(t = 6\,\text{s}\).
The line meets the time axis where \(8 - 2t = 0\), at \(t = 4\,\text{s}\). Above the axis, from \(0\) to \(4\,\text{s}\), is a triangle \(\tfrac12 \times 4 \times 8\). Below it, from \(4\) to \(6\,\text{s}\), is 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 \(20\,\text{m}\).
The engineer at the Metro depot reads her graph the same way. The slopes of the first and last pieces give the acceleration and the braking, \(1\,\text{m/s}^2\) each. The height of the flat piece gives the top speed, \(20\,\text{m/s}\). The whole area gives the distance between the stations, \(1200\,\text{m}\).
A slope must be read in the units of the axes, never with a protractor. The angle a line makes on paper depends on the scales the drawer chose, so \(\tan\) of that angle means nothing by itself. An area must be measured down to the line \(y = 0\). If the vertical axis starts at \(10\,\text{m/s}\) instead of zero, the area inside the drawn box misses a whole rectangle.
A v-t graph is flat at \(15\,\text{m/s}\) for \(4\,\text{s}\), but its vertical axis starts at \(10\,\text{m/s}\). A student measures the area between the line and the bottom of the drawn axes. What does she get, and what is the true displacement?
Graphical Understanding of Motion reads velocity and acceleration as slopes and displacement as an area, and sketches all three graphs from any one. Angular motion is read the same way. Potential Energy and Conservation of Energy reads the force as minus the slope of an energy curve, and equilibrium at its turning points. The work of a gas is the area under a p-V path, and the heating curve of water shows slopes and flat plateaus. Potential and field graphs, and the characteristic curves of conductors and diodes, are read by slope too.
- Reading a slope as the angle of the line on the paper.Compute rise over run in the units of the axes. The drawn angle depends on the scales and has no physical meaning.
- Measuring an area to the bottom of the drawn box.An area is measured to the line \(y = 0\). Check where the vertical axis starts before you count squares.
- Adding every area as positive for the displacement.Area below the axis is negative. Keep the signs for the displacement; drop them only for the distance.