Motion Graphs

A picture that already contains the algebra

In Motion in One Dimension you learned the constant-acceleration formulas, and you used them by picking the one with the right letters in it. That works right up until the acceleration stops being constant. A cart that gets pushed, then coasts, then gets braked has three different accelerations in eight seconds, and no single formula covers it.

A graph covers it. Draw acceleration against time, or velocity against time, or position against time, and everything you want to know is sitting there as a steepness or as an amount of shaded space. The formulas you already know are just the areas of triangles. Once you can see that, a problem that would take three separate applications of $x = x_0 + v_0t + \tfrac12 at^2$ becomes a bit of arithmetic with rectangles.

There are only two questions you ever ask a motion graph. How steep is it here? and how much area is under it so far? This lesson is those two questions, repeated until they are automatic.

Slope takes you down the chain

Start with position. Call it $x$, measured in metres from some origin you chose, and plot it against time $t$. If the object is at $x = 5$ m at $t = 0$ and at $x = 29$ m at $t = 8$ s, it covered 24 m in 8 s. That rate, 24 m divided by 8 s, is 3 m/s, and it is exactly the slope of the line joining those two points, rise over run.

That is not a coincidence or an analogy. Velocity is defined as the rate at which position changes, and slope is what "rate of change" looks like on a graph. So

$v = \frac{\Delta x}{\Delta t} \quad \text{over an interval,} \qquad v = \frac{dx}{dt} \quad \text{at an instant.}$

The first version is the slope of the straight line between two points on the curve, which gives you the average velocity over that stretch. The second is the slope of the tangent, the steepness right at one instant, which is the instantaneous velocity. If the graph is a straight line the two are the same number, because a straight line has the same steepness everywhere.

Acceleration is the same move again, one level down. It is the rate at which velocity changes, so it is the slope of the $v$-$t$ graph:

$a = \frac{\Delta v}{\Delta t}.$

A $v$-$t$ line climbing from 0 to 12 m/s in 4.0 s has slope $12/4.0 = 3.0$ m/s², and that is the acceleration. Units come out right automatically: metres per second, divided by seconds, is metres per second squared.

Area takes you back up the chain

Now run the chain the other way. If an object moves at a steady 6 m/s for 7 s, its displacement is $6 \times 7 = 42$ m. On a $v$-$t$ graph that is a flat line at 6 m/s, and the rectangle beneath it is 6 units tall and 7 units wide. Its area is 42. The area under the velocity graph is the displacement.

That argument only used a constant velocity, but it generalises immediately. Chop any velocity graph into slices so thin that $v$ is nearly constant across each slice. Each slice contributes (its height) $\times$ (its width), a tiny displacement. Add them all up and you have the total. That sum of thin rectangles is what the integral sign means:

$\Delta x = \int v \, dt = \text{area under the } v\text{-}t \text{ graph.}$

The same reasoning one level down gives

$\Delta v = \int a \, dt = \text{area under the } a\text{-}t \text{ graph.}$

So an object starting from rest with a constant $2.5$ m/s² for 6.0 s has velocity equal to the rectangle $2.5 \times 6.0 = 15$ m/s. And a car starting from rest with constant acceleration $a$ has a $v$-$t$ graph that is a straight line from the origin, reaching height $at$ at time $t$. The area under it is a triangle: $\tfrac12 \times \text{base} \times \text{height} = \tfrac12 \times t \times at = \tfrac12 at^2$. That is the formula you memorised last lesson. It is the area of a triangle, and nothing more.

Signed area, and why distance is not displacement

Area below the time axis counts as negative. That is not a rule bolted on; it follows from what the axis means. Below the axis, $v$ is negative, the object is moving backwards, and its displacement over that stretch is negative. A rectangle of "negative height" has negative area.

This is where two words split apart that you may have been using interchangeably. Displacement is the signed area: forwards minus backwards, the net change in position. Distance travelled is the total area counted as positive, forwards plus backwards, what the odometer reads. A ball thrown straight up and caught again has zero displacement and a very nonzero distance.

So whenever a $v$-$t$ line crosses the axis, you must handle the two pieces separately. Find the crossing time first: that is the instant the object stops and turns round. It is also, for the same reason, the instant of maximum displacement: up to then you were adding area, after it you start subtracting.

Three graphs of the same motion

Everything so far links neighbouring graphs. Put all three above each other and the links become visible at once. Take a cart that starts from rest at the origin, is pushed at $+4.0$ m/s² for 2.0 s, coasts with zero acceleration until 5.0 s, then is braked at $-4.0$ m/s² until 8.0 s.

The $a$-$t$ graph is three flat steps. Integrating it (that is, accumulating area) gives a $v$-$t$ graph that ramps up to $4.0 \times 2.0 = 8.0$ m/s, holds flat at 8.0 m/s, then ramps back down to zero. Integrating that gives an $x$-$t$ graph that curves upward, then runs straight, then flattens out.

Notice the smoothing. Where $a$ jumps, $v$ has a kink but no jump. Where $v$ has a kink, $x$ stays perfectly smooth. Each trip up the chain makes the curve one degree gentler, so a discontinuous acceleration can never produce a broken position graph. If you sketch one with a jump in it, you have made a mistake.

Worked examples

1. Ramp and hold. A golf cart's velocity rises linearly from 0 to 12 m/s over the first 4.0 s, then stays at 12 m/s until $t = 10$ s. Find the acceleration over the first stretch, and the total displacement.

The acceleration is the slope of the first piece: $(12 - 0)/(4.0 - 0) = 3.0$ m/s². The displacement is the whole area, and the shape is a triangle glued to a rectangle. Triangle: $\tfrac12 \times 4.0 \times 12 = 24$ m. Rectangle: the flat stretch runs from 4.0 s to 10 s, so its width is 6.0 s, and its area is $6.0 \times 12 = 72$ m. Total: $24 + 72 = 96$ m.

The general shape here (ramp up, hold, ramp down) is a trapezoid, and it is the single most common figure in this topic. Split it into a triangle, a rectangle and a triangle every time.

2. A line that crosses the axis. A velocity graph is a straight line from $+12$ m/s at $t = 0$ down to $-6.0$ m/s at $t = 9.0$ s. Find the maximum displacement from the start, and the displacement over the full 9.0 s.

First the slope: $a = (-6.0 - 12)/(9.0 - 0) = -18/9.0 = -2.0$ m/s². Now find the crossing. Starting at 12 m/s and losing 2.0 m/s every second, the velocity reaches zero at $t = 12/2.0 = 6.0$ s. That is the turning point, so the maximum displacement is the area of the first triangle:

$\tfrac12 \times 6.0 \times 12 = 36 \text{ m}.$

After that the object goes backwards. The second triangle runs from 6.0 s to 9.0 s, a base of 3.0 s, and drops to a depth of $-6.0$ m/s, so its signed area is $\tfrac12 \times 3.0 \times (-6.0) = -9.0$ m. The net displacement is $36 - 9.0 = 27$ m. The distance travelled, if you were asked for it, would be $36 + 9.0 = 45$ m instead.

3. Working backwards to a time. A car's velocity rises linearly from 0 to 30 m/s over $0 \le t \le 6.0$ s, then falls linearly to zero at $t = 10.0$ s. At what time has it travelled exactly 100 m?

Check the first piece before assuming anything: its area is $\tfrac12 \times 6.0 \times 30 = 90$ m. That is short of 100 m, so the answer lies in the second piece, and you need 10 m more.

On the second piece the velocity falls from 30 m/s to 0 over 4.0 s, so its slope is $-30/4.0 = -7.5$ m/s². Measure time from the 6.0 s mark with a new variable $\tau = t - 6.0$. Then $v = 30 - 7.5\tau$, and the extra area is the standard constant-acceleration result:

$\Delta x = 30\tau - \tfrac12(7.5)\tau^2 = 30\tau - 3.75\tau^2 = 10.$

Rearranged, $3.75\tau^2 - 30\tau + 10 = 0$. The quadratic formula gives

$\tau = \frac{30 \pm \sqrt{900 - 4(3.75)(10)}}{2(3.75)} = \frac{30 \pm \sqrt{750}}{7.5} = \frac{30 \pm 27.386}{7.5}.$

The plus sign gives $\tau = 7.65$ s, which is past the end of the second piece and therefore not real. The minus sign gives $\tau = 2.614/7.5 = 0.349$ s. So $t = 6.0 + 0.349 = 6.35$ s.

4. A curved acceleration graph. A car starts from rest at the origin. Its acceleration rises linearly from 0 to 6.0 m/s² over $0 \le t \le 3.0$ s, then falls linearly back to zero at $t = 6.0$ s. How far has it gone at 6.0 s?

Do it one level at a time. On the first piece $a = 2.0t$, since it must reach 6.0 at $t = 3.0$. The velocity is the accumulated area under that line, which up to time $t$ is a triangle of base $t$ and height $2.0t$:

$v = \tfrac12 \times t \times 2.0t = t^2.$

So $v = 9.0$ m/s at $t = 3.0$ s. To get position you need the area under $v = t^2$, and here you need one fact from calculus: the area under $t^n$ up to time $t$ is $t^{n+1}/(n+1)$. It is the same rule that turned the straight line $v = at$ into $\tfrac12 at^2$. With $n = 2$:

$x(3.0) = \frac{(3.0)^3}{3} = 9.0 \text{ m}.$

For the second piece, write $a = 12 - 2.0t$, which is 6.0 at $t = 3.0$ and zero at $t = 6.0$. Accumulating its area from 3.0 s onward and adding the 9.0 m/s you already had gives $v = -18 + 12t - t^2$. Check it at $t = 6.0$: $-18 + 72 - 36 = 18$ m/s. That matches the total area of the acceleration triangle, $\tfrac12 \times 6.0 \times 6.0 = 18$ m/s, which is a good sign that nothing went wrong.

Accumulating area under that velocity from 3.0 s to 6.0 s gives

$\left[-18t + 6t^2 - \tfrac{t^3}{3}\right]_{3.0}^{6.0} = (-108 + 216 - 72) - (-54 + 54 - 9) = 36 + 9 = 45 \text{ m}.$

Total: $9.0 + 45 = 54$ m.

Traps that cost real points

Reading height as slope. Covered above, and it is the single most common error. The top of a hill on an $x$-$t$ graph is where $v = 0$, not where $v$ is biggest.

Forgetting the width of the middle piece. In example 1 the flat stretch runs from 4.0 s to 10 s, so its width is 6.0 s, not 10 s. Trapezoids punish carelessness here more than anything else.

Averaging velocities instead of using area. Average velocity is displacement divided by total time, not the average of the numbers you see on the graph. For a staircase graph that sits at 5, 10, 15 and 20 m/s for 2 s each those happen to agree, both giving 12.5 m/s, because the intervals are equal. For the trapezoid in Figure 3 they do not agree at all. Compute the area, then divide by the total time.

Mixing distance into a displacement question, or the reverse. Read the last five words of the question again before you write the number down.

How to recognise this technique

Reach for slopes and areas whenever the problem hands you a graph instead of a list of numbers, or describes motion in pieces: "for the first 3 seconds... then... then". The word "then" is the signal. A single formula cannot span a "then"; a sum of areas can.

Three specific triggers. If the question gives a velocity graph and asks for displacement, distance or average velocity, you want area. If it gives a velocity or position graph and asks for acceleration or velocity, you want slope. If it asks for a maximum displacement or the moment something turns round, find where $v$ crosses zero and take the area up to there.

And if the shape it describes is a straight line, a triangle, a rectangle or a trapezoid, the answer is arithmetic on areas you learned in primary school. Sketch the graph even when the problem does not ask you to. A five-second sketch turns most of these into one line of working.

Next comes the most common constant acceleration in physics, and the complications that arrive when acceleration refuses to stay constant: Free Fall and Variable Acceleration.