Position, velocity and acceleration graphs

A curve of position against time tells us where an object is, which way it is moving and how its velocity changes. The trick is to read the height of the curve and its slope separately.

Picture a cart on a straight track. It moves to the right, slows down, turns round and comes back to the left. Saying only that the cart ended where it started throws away most of what happened. We want a way of talking that accounts for the whole motion.

This chapter deals only with motion along one straight line, and a fixed point on the cart stands for its position. We do not yet ask why the cart moves as it does. Derivatives and integrals are introduced from the graphs as we go; you do not need to have met them before.

1.1 Say where you measure from

Choose an origin on the track and take the direction to the right as positive. The cart's position is $x$ , in metres; the time is $t$ , in seconds.

When the cart is three metres to the right of the origin, its position is $x=3\,\mathrm m$ . Three metres to the left, it is $x=-3\,\mathrm m$ . The sign says which side of the origin the cart is on.

A position on its own does not tell us which way the cart is going. Passing the three-metre mark, it may still be heading right or already be on its way back. To find the direction of motion we have to compare positions at neighbouring times.

If a time interval starts at $t_1$ and ends at $t_2$ , the displacement is the final position minus the initial one:

\Delta x=x(t_2)-x(t_1).

The symbol $\Delta$ means the change between the two times. A displacement can be positive, negative or zero.

Distance travelled counts every stretch the cart covers. If the cart goes four metres to the right and then comes back along the same track, it has travelled eight metres and its displacement is zero. These are the usual one-dimensional definitions of position, displacement and distance, as in OpenStax University Physics Volume 1, Section 3.1.

1.2 Drawing a round trip as a curve

Here is a motion set up for this chapter. The numbers in the table come from a given mathematical model. They are ideal positions, not camera data.

**Table 1.1**　Positions in a round trip at five times.
Time (s)	Position (m)
0	0
1	3
2	4
3	3
4	0

To say what happens between the listed times, we also specify that from zero to four seconds the position follows

x(t)=ut-bt^2, \qquad u=4\,\mathrm{m\,s^{-1}}, \qquad b=1\,\mathrm{m\,s^{-2}}.

Here $u$ and $b$ are given constants. Both terms are lengths, so they can be subtracted. You do not need derivatives yet: substitute each time and you can check every position in the table.

Figure 1.1　Position (top), velocity (middle) and acceleration (bottom), all against time. The dashed line marks the turn at two seconds. The two shaded areas in the velocity graph have signed areas of plus four metres and minus four metres.

All three curves are computed from the ideal motion given in this chapter. The dots are the positions in Table 1.1; they are not extra measurements.

Look first at the top graph. For the first two seconds the position keeps increasing; for the last two it keeps decreasing. The curve rises and then falls as the cart reaches its furthest point to the right.

The horizontal axis is time and the vertical axis is position, so the bend on the page is not the shape of the track. The cart stays on its straight track the whole time.

The five positions in the table are not enough to fix the whole motion. The cart could go a little further between two listed times and come back. We can draw a definite smooth curve here only because the position function is given as well. The next chapter asks how far we can get from a discrete record alone.

1.3 Velocity is the slope of the position graph

Between zero and one second the cart moves three metres to the right. Its average velocity over that second is three metres per second. In general:

\overline v =\frac{\Delta x}{\Delta t} =\frac{x(t_2)-x(t_1)}{t_2-t_1}, \qquad t_2>t_1.

On the position graph, the straight line joining the two times has exactly this slope. The vertical change is in metres and the horizontal change in seconds, so the slope is in metres per second.

The cart's motion is not uniform, so an average over a whole second need not equal the velocity at one instant. To ask how fast it is going at one second, compare one second with one and a half seconds, then with one point one seconds, and keep shrinking the interval.

With this chapter's position function,

\begin{aligned} \frac{x(t+\Delta t)-x(t)}{\Delta t} &=\frac{u\Delta t-b\bigl(2t\Delta t+(\Delta t)^2\bigr)}{\Delta t}\\ &=u-2bt-b\Delta t. \end{aligned}

Table 1.2 lists the result for $t=1\,\mathrm s$ .

**Table 1.2**　Average velocity over shorter and shorter intervals starting at one second.
Interval after one second (s)	Average velocity over the interval ( $\mathrm{m\,s^{-1}}$ )
0.5	1.5
0.1	1.9
0.01	1.99

As the interval shrinks, the result approaches two metres per second. The limit as the interval goes to zero is the instantaneous velocity, usually just called the velocity. It is also the derivative of the position function with respect to time:

v(t)=\lim_{\Delta t\to0} \frac{x(t+\Delta t)-x(t)}{\Delta t} =\frac{\mathrm dx}{\mathrm dt}.

On the graph, the line through two points turns into the tangent at the point. The slope of the tangent is the rate at which position changes at that instant. Compare the definitions of instantaneous velocity and speed in OpenStax Volume 1, Section 3.2.

For this chapter's motion, the algebra above has already given

v(t)=u-2bt.

So the velocity is plus two metres per second at one second and minus two metres per second at three seconds. The cart passes the same position twice, moving in opposite directions.

In one dimension the sign of the velocity gives the direction. Speed is the size of the velocity, written $\lvert v\rvert$ . At both of those times the speed is two metres per second.

When you read a position graph, ask in turn: is the curve rising or falling? How steep is it? Is the steepness growing or shrinking? The height of the curve answers only the question of position. The other questions are about velocity.

1.4 Acceleration says how velocity changes

The velocity graph has a slope too. It measures how quickly the velocity changes and is called the acceleration:

\overline a=\frac{\Delta v}{\Delta t}, \qquad a(t)=\frac{\mathrm dv}{\mathrm dt}.

Acceleration is measured in metres per second squared: how many metres per second the velocity changes by in each second. It keeps a direction as well; see OpenStax Volume 1, Section 3.3.

In this chapter's motion the velocity drops by two metres per second every second, so the acceleration is constant throughout:

a=-2b=-2\,\mathrm{m\,s^{-2}}.

In the first two seconds the cart moves right and its velocity is positive. The negative acceleration brings the velocity towards zero, so the cart slows down. In the last two seconds the cart moves left and its velocity is negative. The velocity keeps decreasing, so its size grows and the cart speeds up.

A negative acceleration therefore does not always mean slowing down. For motion along a line with non-zero velocity, the speed increases when velocity and acceleration point the same way and decreases when they point opposite ways.

At the instant of the turn the velocity is zero but the acceleration is still minus two metres per second squared. The cart stops only for an instant and then moves off to the left. To decide whether it is at rest over some time, check whether its position stays fixed over that time.

1.5 Getting displacement back from the velocity graph

If the velocity is constant, the displacement is the velocity times the time elapsed. On the velocity graph that is the signed area of a rectangle.

When the velocity changes, split the time into many short pieces, estimate the displacement in each piece from a representative velocity, and add them up. As the pieces get shorter, the sum becomes the integral of velocity over time:

x(t_2)-x(t_1)=\int_{t_1}^{t_2}v(t)\,\mathrm dt.

For now, read the integral as a continuous running total. Area below the time axis counts as negative, because during that part of the motion the position decreases.

In Figure 1.1 the first two seconds form a triangle with a base of two seconds and a height of four metres per second, so its area is four metres. The triangle for the last two seconds lies below the axis, and its signed area is minus four metres. Together they give a displacement of zero.

For the distance travelled, the return trip adds length too. Add the sizes of the two stretches and you get eight metres. Equivalently,

D=\int_0^{4\,\mathrm s}\lvert v(t)\rvert\,\mathrm dt=8\,\mathrm m.

The velocity graph fixes how much the position changes. To know where the cart actually is, we also need its starting position:

x(t)=x(0)+\int_0^t v(s)\,\mathrm ds.

The $s$ inside the integral is only a variable that runs over time; it is still measured in seconds. Two carts can have identical velocity graphs and stay two metres apart the whole time. That gap has to come from the initial position.

1.6 Worked example: a motion that returns to its start

Take the motion of Figure 1.1 from zero to four seconds. Find the time of the turn, the total displacement, the total distance, the average velocity and the average speed.

First decide how to split the motion. Before and after the turn the cart moves in different directions, so for the distance we split at the turn. The velocity relation gives the time of the turn:

u-2bt=0 \quad\Longrightarrow\quad t=\frac{u}{2b}=2\,\mathrm s.

Substituting into the position function gives the furthest point to the right, four metres. The cart goes from zero to four metres and back from four metres to zero, so

\begin{aligned} \Delta x&=0\,\mathrm m,\\ D&=4\,\mathrm m+4\,\mathrm m=8\,\mathrm m. \end{aligned}

Average velocity is displacement divided by total time; average speed is total distance divided by total time:

\overline v=\frac{0\,\mathrm m}{4\,\mathrm s}=0\,\mathrm{m\,s^{-1}}, \qquad \overline{\lvert v\rvert} =\frac{8\,\mathrm m}{4\,\mathrm s} =2\,\mathrm{m\,s^{-1}}.

The average velocity is zero, which fits the cart ending where it began. The average speed is positive, which fits the eight metres it actually covered. The two signed areas on the velocity graph cancel, an independent check on the displacement.

1.7 Comparing in the Motion Graph Studio

Open the Motion Graph Studio. You can specify a position, velocity or acceleration curve there and see the other two curves together with the cart.

Start with uniform motion. Choose “Uniform motion” and drag the gold clock. Before you do, sketch the position, velocity and acceleration graphs you expect. Afterwards, say which graph is a sloping straight line and which is horizontal, and what the height of each horizontal line means.

Then look at a return. Choose “Reverse and return”. From the velocity curve, predict when the cart moves right and when it moves left, then drag the clock to check. Move the clock to the end and note the “Signed velocity area” and “Travelled distance” readouts. Explain why the two readings can differ.

Finally, change the start. Choose “Uniform motion” again and note the initial position. Leave the velocity curve alone, raise “Integration initial position” by two metres, and compare the whole position graph, the velocity graph, the displacement and the distance. Say in one sentence which information has to come from the initial value.

If you are not running the studio, Table 1.3 lets you follow the argument. The clock is at the end of the record in every case.

**Table 1.3**　Motion Graph Studio results at the end of the record for three settings.
Setting	Duration (s)	Final position (m)	Signed velocity area (m)	Travelled distance (m)
Reverse and return (default)	6	0	0	8.8
Uniform motion	6	6	6	6
Uniform motion, integration initial position 2 m	6	8	6	6

In Reverse and return the cart reaches its furthest point, 4.4 metres, at three seconds and then returns to the origin. Its displacement is zero and its distance is 8.8 metres. Changing the integration initial position shifts the whole position graph up by two metres; the velocity graph, the displacement and the distance stay the same.

The studio's curves are smooth interpolations through seven knots, and the cart can turn between knots, so look at the whole curve. The studio prescribes an ideal motion and does not compute the forces needed to produce it; “Export model CSV” saves model values. The worked example of Figure 1.1 is a separate motion set up for this chapter, with its own parameters.

1.8 Exercises

A car is three metres to the right of the origin. From this position alone, can you tell whether it is moving left or right? What else would you need?
A cart goes from two metres left of the origin to three metres right of it, always moving right, in five seconds. Find the displacement, the distance, the average velocity and the average speed.
Over some time interval a position graph lies above the time axis but slopes downward throughout. What are the signs of the position and the velocity? Is the cart getting closer to the origin?
For the motion of Figure 1.1, consider only one to three seconds. Find the displacement, distance, average velocity and average speed over that interval.
At some instant a cart has a velocity of minus two metres per second and an acceleration of plus one metre per second squared. Right around that instant, is its speed increasing or decreasing? If the acceleration stays the same, what happens two seconds later?
Two carts have identical velocity graphs and start two metres apart. How are their position graphs, displacements and distances related?
A camera records the position once a second, and the five positions are exactly those of Table 1.1. Can you conclude that the instantaneous velocity at two seconds must be zero? What assumption does your answer depend on?
Design a motion that goes right, then left, and ends at its start. Give a constant velocity and a duration for each part, and compute the average velocity and average speed. What does this piecewise model idealise at the turn?
Transfer. A lift rises 30 metres from the ground in 15 seconds, waits at the floor for 10 seconds, then goes down 12 metres in 8 seconds. Take upward as positive and treat each part as uniform motion. Sketch the velocity graph and find the displacement, distance, average velocity and average speed for the whole trip. What does this approximation leave out when the lift starts and stops?

1.9 Answers and explanations

Exercise 1

No. Position only says where the car is. To tell the direction you need the change in position between neighbouring times, or the velocity at that time.

Exercise 2

The displacement is plus five metres, and since the cart always moves right the distance is also five metres. The average velocity is plus one metre per second and the average speed is one metre per second. If the cart had doubled back on the way, the distance could be larger while the final position and the displacement stayed the same.

Exercise 3

The position is positive and the velocity negative. As long as the curve stays above the axis, the cart is approaching the origin from the right. If it carries on left past the origin, it moves away from the origin again.

Exercise 4

The positions at one and three seconds are both three metres, so the displacement and the average velocity are zero. The cart goes from three metres to four and back to three, a distance of two metres. Over two seconds that is an average speed of one metre per second.

Exercise 5

Velocity and acceleration point in opposite directions, so the speed is decreasing. If the acceleration stays the same, the velocity reaches zero after two seconds; with the same acceleration the cart then moves to the right. Being at rest for an instant at two seconds does not mean it stays at rest.

Exercise 6

The position graphs are always two metres apart vertically. Over any time interval the two carts change position by the same amount and have the same speed at every instant, so their displacements are equal and so are their distances.

Exercise 7

No. Five records do not fix the motion between them. With the quadratic position function given in this chapter the velocity at two seconds is zero. With discrete positions alone, you can build smooth curves through the same five points that have a different slope at two seconds.

For example, take $x(t)=ut-bt^2+k\sin(\pi t/\tau)$ with $\tau=1\,\mathrm s$ and any length $k$ . The sine term vanishes at whole seconds, so the five positions are unchanged, but the velocity at two seconds becomes $k\pi/\tau$ . To pick out a single curve you must also assume the motion is smooth enough, or record more often.

Exercise 8

For example, move at plus two metres per second for two seconds, then at minus one metre per second for four seconds. The displacement is plus four metres plus minus four metres, which is zero. The distance is eight metres in six seconds, so the average speed is four thirds of a metre per second and the average velocity is zero.

The velocity jumps at the turn, so the ordinary instantaneous acceleration is not defined there. A real cart needs a finite time to change its velocity.

Exercise 9

The velocity graph is three horizontal pieces: plus two metres per second from zero to fifteen seconds, zero from fifteen to twenty-five seconds, and minus one and a half metres per second from twenty-five to thirty-three seconds.

The displacement is $30\,\mathrm m-12\,\mathrm m=18\,\mathrm m$ and the distance $30\,\mathrm m+12\,\mathrm m=42\,\mathrm m$ , over 33 seconds. The average velocity is about $0.55\,\mathrm{m\,s^{-1}}$ upward and the average speed about $1.27\,\mathrm{m\,s^{-1}}$ . The ten-second wait adds no displacement but does count in the total time.

The uniform approximation makes the velocity jump when the lift starts and stops, as if the acceleration were infinite. A real lift takes a few seconds to speed up and slow down. As long as each part keeps the same displacement, the displacement, distance and averages are unchanged, but the corners of the velocity graph become ramps.

Next: what does a camera actually record?

This chapter started from a known position curve, found velocity from its slope and displacement from the signed area under the velocity graph. A real camera gives positions only at a finite set of times. The next chapter starts from such a record and looks at how calibration, frame rate and position noise affect what we can say about the motion.

Figure and data notes

The basic definitions follow OpenStax University Physics Volume 1, Sections 3.1, 3.2 and 3.3. The worked example, the exercises and their numbers were written for this chapter.
Figure 1.1 is drawn from this chapter's position function and its exact derivatives over zero to four seconds, with dots one second apart. No measured data are used.
Table 1.3 was computed with the Motion Graph Studio model code at the default duration, with the clock at the end of the record. The model has no random element, so a rerun gives the same numbers.

Test it in the Lab

Change one setting at a time, then compare what you see with the tables and answers in this chapter.

Motion Graph Studio →Draw the position, velocity or acceleration curve and watch a 3D cart follow it. Slopes and areas, with their signs, show when the cart turns back and why distance and displacement differ.

Sources and reading scope

OpenStax, University Physics Volume 1, §3.1 Position, Displacement, and Average Velocity
Definitions of position, displacement, distance travelled and average velocity in one dimension. The chapter’s examples and exercises are written independently.
OpenStax, University Physics Volume 1, §3.2 Instantaneous Velocity and Speed
Instantaneous velocity as the limit of average velocity and as the derivative of position; speed as its magnitude.
OpenStax, University Physics Volume 1, §3.3 Average and Instantaneous Acceleration
Average and instantaneous acceleration, their units and sign in one dimension.