Motion graphs and their interpretation

A motion graph plots a kinematic quantity against time, and the physics is carried by the shape of the curve, not by a picture of the path. On a position–time graph $x(t)$, the slope of a segment between two instants gives the average velocity $v_{\mathrm{avg}}=\frac{\Delta x}{\Delta t}$, while the slope of the tangent at one instant gives the instantaneous velocity. A horizontal line therefore means rest, a slanted straight line means motion at constant velocity, and a curve means motion with changing velocity, that is, with acceleration. On a velocity–time graph $v(t)$, the slope gives acceleration $a=\frac{\Delta v}{\Delta t}$, and the area between the graph and the time axis gives displacement $\Delta x=\sum v\,\Delta t$. Areas above the axis count as positive and areas below it as negative, so after a reversal they partly cancel and the displacement is smaller than the distance travelled. Always read the axis labels and units first: confusing $x(t)$ with $v(t)$ is the most common error in graph reading.

A pedestrian walks along a straight street. The $x(t)$ graph is a broken line through the points $(0\ \text{s}, 0\ \text{m})$, $(4\ \text{s}, 12\ \text{m})$, $(10\ \text{s}, 12\ \text{m})$, $(14\ \text{s}, 0\ \text{m})$. Find the average velocity on each of the three intervals and describe the pedestrian's motion.

A car's velocity–time graph $v(t)$ is a broken line: during the first $4\ \text{s}$ the velocity rises uniformly from $0$ to $8\ \text{m/s}$, for the next $4\ \text{s}$ it stays at $8\ \text{m/s}$, and during the last $4\ \text{s}$ it falls uniformly to $0$. Find the acceleration in each stage and the total displacement.

A trolley moves along a straight rail. Its $v(t)$ graph is: $+6\ \text{m/s}$ from $t=0$ to $3\ \text{s}$, then $-2\ \text{m/s}$ from $3\ \text{s}$ to $5\ \text{s}$, then $+1\ \text{m/s}$ from $5\ \text{s}$ to $9\ \text{s}$. (a) Find the displacement and the distance travelled. (b) Find the average velocity and the average speed over the whole $9\ \text{s}$. (c) Explain why these two averages differ.

A 30–40 minute self-study plan, carried out entirely in a simulation.

1. Open a motion simulation such as PhET The Moving Man, or any other tool that plots $x(t)$ and $v(t)$ graphs. Keep the motion one-dimensional.
2. Generate three motions and record them: (a) the body is at rest; (b) motion at constant velocity $+2\ \text{m/s}$; (c) uniformly accelerated motion from rest. For each one, pause and examine both graphs.
3. For case (b), pick two points on the $x(t)$ graph, compute the slope by hand as $\frac{\Delta x}{\Delta t}$ and compare it with the velocity readout. For case (c), take two instants, compute $\frac{\Delta v}{\Delta t}$ and compare it with the acceleration readout.
4. Then hide the $v(t)$ graph and sketch from the $x(t)$ graph what the velocity graph must look like; afterwards reveal it and correct your sketch.

Safety. Do all measurements in the simulation. Do not time your own motion on roads, stairs or anywhere near traffic.

Show that one and the same motion is encoded in two ways: the slope of the $x(t)$ graph equals the value on the $v(t)$ graph, and the area under $v(t)$ equals the change in $x$.

A sketch of two graphs stacked vertically and sharing one horizontal time axis. On top, $x(t)$ is a slanted straight line with a marked right triangle and the label $\frac{\Delta x}{\Delta t}$ along its hypotenuse. Below, $v(t)$ is a horizontal straight line, with the region between it and the time axis shaded and labelled $\Delta x = v\,\Delta t$. A dashed vertical line links the two graphs at one and the same instant.

In a school 100 m race the stopwatch is read at five marks. The data are: $x=0$ m at $t=0$ s, $x=12$ m at $t=2.0$ s, $x=26$ m at $t=4.0$ s, $x=42$ m at $t=6.0$ s, $x=60$ m at $t=8.0$ s, $x=78$ m at $t=10.0$ s and $x=100$ m at $t=12.5$ s. Plotting $x$ vertically against $t$ horizontally gives a curve that is shallow at first and then steepens: the runner covers $12$ m in the first two seconds, $18$ m in $2.0$ s in the middle of the race, and $22$ m in the last $2.5$ s.

The central reading rule is used here: the slope of a segment equals $\Delta x/\Delta t$. For the first $2.0$ s it is $12/2.0 = 6.0$ m/s, for the middle part $18/2.0 = 9.0$ m/s, for the last one $22/2.5 = 8.8$ m/s. The straight line joining $(0;0)$ and $(12.5;100)$ has slope $100/12.5 = 8.0$ m/s — the average speed of the whole race — while the steepest part of the curve marks the moment of greatest speed.

A city bus pulls away from a stop on a straight avenue: its speed rises from zero to $20$ m/s in $8.0$ s, holds at $20$ m/s for $60$ s, then drops to zero in $5.0$ s at the traffic light. The on-board logger plots speed against time.

The distance travelled equals the area between the speed line and the time axis: $\tfrac{1}{2}\cdot 20\cdot 8.0 + 20\cdot 60 + \tfrac{1}{2}\cdot 20\cdot 5.0 = 80 + 1200 + 50 = 1330\ \text{m} \approx 1.3\ \text{km}.$ The slopes of the rising and falling pieces of the same graph give the accelerations $20/8.0 = 2.5\ \text{m/s}^2$ and $-20/5.0 = -4.0\ \text{m/s}^2$.

Weather radar tracks the leading edge of a storm line moving east towards a town. The fixes are: $30$ km west of the town at $15{:}00$, $18$ km west at $16{:}00$, $6$ km west at $17{:}00$, and the town itself ($0$ km) at $19{:}00$. Taking east as positive, $x=-30$ km at $15{:}00$, $x=-18$ km at $16{:}00$, $x=-6$ km at $17{:}00$ and $x=0$ km at $19{:}00$.

The first three points lie on a single straight line, and its slope gives the front's velocity: $24$ km in $2.0$ h, that is $12$ km/h $\approx 3.3$ m/s eastward. The last segment runs from $x=-6$ km to $x=0$ km in $2.0$ h, with a slope of only $3$ km/h. The kink in the graph at $17{:}00$ is direct evidence that the motion stopped being uniform and the front slowed down, even though nobody measured its speed directly.