Position, velocity, acceleration and displacement

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Physics · Space, time and motion · SL · Guide 2025

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Position, velocity, acceleration and displacement

Kinematics describes motion without asking what causes it.

Three quantities do the describing: position, velocity and acceleration.

Each one is found from the one before it by asking how fast it changes, and each can be recovered from the one after it by adding up the changes.

Position and displacement

Position is where a body is, measured from a chosen origin along a chosen positive direction.

In one dimension a single signed number does the job: $x = -3.0$ m means 3.0 m on the negative side of the origin.

You choose the origin and the positive direction, and you must then keep them for the whole problem.

Displacement is the change in position, $\Delta x = x_2 - x_1$ .

It is a vector: it has a size and a direction, shown by its sign in one dimension.

A body that ends where it started has zero displacement, however far it moved in between.

Position depends on where you put the origin; displacement does not.

Move the origin 5 m to the left and every position grows by 5 m, but every difference between two positions stays the same.

That is why the equations of motion are written in terms of displacement, not position.

Rates of change

Velocity is the rate of change of position, $v = \dfrac{\Delta x}{\Delta t}$ .

A velocity of $-2.0$ m s⁻¹ means the position decreases by 2.0 m every second: the body moves in the negative direction at 2.0 m s⁻¹.

Acceleration is the rate of change of velocity, $a = \dfrac{\Delta v}{\Delta t}$ , measured in m s⁻².

An acceleration of 3.0 m s⁻² means the velocity rises by 3.0 m s⁻¹ every second.

Both velocity and acceleration are vectors, so both carry a sign.

The chain is the same at each link: position changes at a rate called velocity, and velocity changes at a rate called acceleration.

A body can have zero velocity and a non-zero acceleration at the same instant.

A ball at the top of a vertical throw is momentarily at rest, but its velocity is still changing, so its acceleration is not zero.

Speeding up or slowing down

The sign of the acceleration alone does not tell you whether a body is speeding up.

Compare it with the sign of the velocity.

Velocity

Acceleration

What happens

positive

moves in the positive direction, speeding up

positive

negative

moves in the positive direction, slowing down

negative

moves in the negative direction, speeding up

negative

positive

moves in the negative direction, slowing down

Same signs: speeding up.

Opposite signs: slowing down.

A car reversing (negative velocity) while braking has a positive acceleration.

"Deceleration" is an everyday word for an acceleration that opposes the velocity; in a calculation give the acceleration with its sign.

The three graphs

Each quantity can be plotted against time, and the graphs are linked by gradients and areas.

Work out what a gradient or area means from the units of the axes: metres per second is a velocity.

Position-time (x-t) graph: the gradient is the velocity. A straight line means constant velocity; a horizontal line means at rest; a curve that steepens means speeding up.

Velocity-time (v-t) graph: the gradient is the acceleration. The area between the line and the time axis is the displacement, counted negative where the line is below the axis.

Acceleration-time (a-t) graph: the area under it is the change in velocity.

Going down the chain (x to v to a) you take gradients.

Going back up (a to v to x) you find areas, and you need a starting value, because an area gives only a change.

Finding gradients at a single instant is covered in the section on average and instantaneous values; areas under curved graphs in the section on non-uniform acceleration.

A real position-time graph has no sharp corners, and a real velocity-time graph no vertical jumps.

A corner on an x-t graph would mean the velocity changed in zero time, which needs an infinite acceleration.

Sharp corners in a question are an idealization of a change that happens quickly.

✓

EXAM FOCUS

How the IB examines this

Paper 1A, definitions and signs: a stem gives a velocity and an acceleration with signs and asks what the body is doing, or which pair of quantities are both vectors. Marks are lost by reading "negative acceleration" as "slowing down".

Paper 1A, graphs: the options are four sketch graphs or a table of "velocity / acceleration" descriptions for a given graph. The skill is reading the gradient of each graph and knowing which graph's gradient is which quantity.

Paper 1B, data: a table of positions from a motion sensor, a video or light gates, with an uncertainty in the heading. You find displacements and velocities over intervals, propagate the uncertainty in a difference, and decide whether the velocity or the acceleration is constant within that uncertainty.

Paper 2, describe: "describe the motion" between two times from a graph. Full marks need the direction, whether the body speeds up, slows down or moves at constant velocity, and a value where the graph gives one.

Paper 2, calculate and sketch: an acceleration from a v-t graph, then a sketch of the matching a-t or v-t graph on labelled axes, often followed by a force from $F = ma$ .

◆

Key terms

Position: where a body is, measured from a chosen origin along a chosen positive direction.

Displacement: the change in position of a body; a vector, with magnitude and direction.

Velocity: the rate of change of position (displacement) with time.

Acceleration: the rate of change of velocity with time.

↗

Worked example 1: positions from a motion sensor

A motion sensor at one end of a bench records the position of a toy car every 0.50 s.

The data below were written for this example.

The time readings can be taken as exact.

t / s

0.50

1.00

1.50

2.00

2.50

3.00

x / m ± 0.005 m

0.200

0.250

0.400

0.650

0.950

1.250

1.550

(a) Calculate the displacement of the car from 0 to 3.00 s. [1]
(b) Calculate the average velocity in each 0.50 s interval. [2]
(c) Determine the acceleration during the first 1.50 s. [2]
(d) Calculate the velocity from 2.00 s to 2.50 s with its uncertainty. [2]

1. (a) $\Delta x = 1.550 - 0.200 = 1.350$ m in the positive direction. [1]

2. (b) $v = \Delta x / 0.50$ for each interval: $0.050/0.50 = 0.10$ m s⁻¹ [1], then 0.30, 0.50, 0.60, 0.60, 0.60 m s⁻¹. [1]

3. (c) An average velocity over an interval is the velocity at the middle of it for uniform acceleration. Over the first three intervals the velocity rises by 0.20 m s⁻¹ every 0.50 s (0.10, 0.30, 0.50 m s⁻¹) [1], so $a = 0.20 / 0.50 = 0.40$ m s⁻². [1]

4. (d) Each position is uncertain by 0.005 m, and the uncertainties of a difference add: $\Delta x = 0.300 \pm 0.010$ m. [1] Then $v = 0.300/0.50 = 0.60$ m s⁻¹; dividing by an exact time divides the absolute uncertainty by the same number, $0.010/0.50 = 0.02$ m s⁻¹, so $v = 0.60 \pm 0.02$ m s⁻¹. [1]

Answer: 1.350 m; 0.10, 0.30, 0.50, 0.60, 0.60, 0.60 m s⁻¹; 0.40 m s⁻²; $0.60 \pm 0.02$ m s⁻¹.

The car accelerates uniformly from rest, then moves at a constant 0.60 m s⁻¹ from 1.50 s.

↗

Worked example 2: a positive acceleration that slows a car down

A car moving in the negative direction changes its velocity from $-15$ m s⁻¹ to $-6.0$ m s⁻¹ in 3.0 s.

Find its acceleration and state whether it is speeding up.

1. $a = \dfrac{v - u}{\Delta t} = \dfrac{-6.0 - (-15)}{3.0}$ [1]

2. $a = +3.0$ m s⁻². [1]

3. The velocity is negative and the acceleration positive: opposite signs, so the car is slowing down. Its speed falls from 15 m s⁻¹ to 6.0 m s⁻¹. [1]

Answer: $+3.0$ m s⁻²; the car is slowing down.

Position against time for a robot: a straight line from 0 m at 0 s to 4.0 m at 8 s, horizontal at 4.0 m until 10 s, then a straight line down to 1.0 m at 16 s

↗

Worked example 3: from a position-time graph to a velocity-time graph

A robot vacuum cleaner moves along a straight corridor.

Its position is shown on the graph above.

Describe its motion and sketch its velocity-time graph.

1. 0 to 8 s: gradient $= 4.0 / 8.0 = +0.50$ m s⁻¹, constant velocity in the positive direction.

2. 8 to 10 s: horizontal line, gradient zero: the robot is at rest at $x = 4.0$ m.

3. 10 to 16 s: gradient $= (1.0 - 4.0)/(16 - 10) = -0.50$ m s⁻¹: it moves back towards the origin at 0.50 m s⁻¹.

4. Velocity-time graph: a horizontal line at $+0.50$ m s⁻¹ from 0 to 8 s, along the time axis from 8 to 10 s, then a horizontal line at $-0.50$ m s⁻¹ from 10 to 16 s.

5. Displacement from 0 to 16 s: $1.0 - 0 = +1.0$ m. Check with the v-t graph: area $= 0.50 \times 8 - 0.50 \times 6 = 4.0 - 3.0 = 1.0$ m.

Answer: constant velocity out, a pause, constant velocity back; the v-t graph is three horizontal segments at $+0.50$ , 0 and $-0.50$ m s⁻¹.

The vertical jumps between them are an idealization: a real robot needs a short time to change velocity.

Velocity against time for a tram: rising in a straight line from 0 to 12 m/s over 20 s, horizontal at 12 m/s until 60 s, then falling in a straight line to zero at 75 s

↗

Worked example 4: a tram between stops (Paper 2 style)

A tram of mass $3.6 \times 10^4$ kg starts from rest at one stop.

It accelerates uniformly to 12 m s⁻¹ in 20 s, travels at 12 m s⁻¹ for 40 s, then slows uniformly to rest at the next stop in 15 s, as in the graph above.

(a) Calculate the acceleration in the first 20 s. [1]
(b) Calculate the acceleration while the tram slows down. [1]
(c) Show that the distance between the stops is about 700 m. [2]
(d) Calculate the magnitude of the resultant force on the tram while it slows down. [1]
(e) Sketch the acceleration-time graph for the journey, with values. [2]

1. (a) $a = 12/20 = 0.60$ m s⁻². [1]

2. (b) $a = (0 - 12)/15 = -0.80$ m s⁻² (opposite to the velocity). [1]

3. (c) Displacement is the area under the v-t graph: $\tfrac{1}{2}(12)(20) + (12)(40) + \tfrac{1}{2}(12)(15)$ [1] $= 120 + 480 + 90 = 690$ m. [1] The full substitution, or the answer to more figures than the 700 m given, earns the marks; working back from 700 m earns at most one.

4. (d) $F = ma = 3.6 \times 10^4 \times 0.80 = 2.9 \times 10^4$ N. [1]

5. (e) Horizontal line at $+0.60$ m s⁻² from 0 to 20 s, zero from 20 to 60 s [1]; horizontal line at $-0.80$ m s⁻² from 60 to 75 s, below the axis [1].

Answer: 0.60 m s⁻²; $-0.80$ m s⁻²; 690 m; $2.9 \times 10^4$ N; three horizontal segments.

Part (d) uses Newton's second law, the usual next step in a kinematics question.

✦

Exam tip

Define acceleration as the rate of change of velocity, not of speed: direction matters, which is why a body going round a bend at constant speed is still accelerating.

If you answer with an equation such as $a = \Delta v / \Delta t$ , say what each symbol stands for, or the definition mark is not awarded.

Common mistakes

Reading a negative acceleration as "slowing down". It means only that the acceleration points in the negative direction; compare its sign with the velocity's.

Treating position and displacement as the same thing. Position is measured from the origin; displacement is a change in position between two moments.

Saying the acceleration is zero when the velocity is zero. At the top of a vertical throw the velocity is zero but it is still changing.

Reading the height of a v-t graph as the position, or the gradient of an x-t graph as the acceleration. Check the units of gradient and area first.

Changing the positive direction halfway through a problem, which flips signs and ruins the final answer.

Check your understanding

1. A body has a velocity of $-4.0$ m s⁻¹ and an acceleration of $+2.0$ m s⁻². Which describes its motion at this instant? [1]
A. It moves in the negative direction and speeds up.
B. It moves in the negative direction and slows down.
C. It moves in the positive direction and speeds up.
D. It moves in the positive direction and slows down.

2. A position-time graph is a straight line with a negative gradient. Which row describes the velocity and the acceleration? [1]

Velocity

Acceleration

negative and constant

zero

negative and constant

negative

negative and decreasing

negative

zero

negative

3. A ball is thrown vertically upwards and caught at the same height. Air resistance is negligible and upwards is positive. Which describes the velocity-time graph from release to catch? [1]
A. A V shape, falling to zero at the top and rising again.
B. A horizontal line below the time axis.
C. A straight line with a constant negative gradient, crossing the time axis halfway through the flight.
D. A straight line with a constant negative gradient that reaches zero at the catch.

4. A body moves in a straight line. Which statements must be true at an instant when its velocity is zero? [1]

I. The gradient of its position-time graph is zero.
II. Its acceleration is zero.
III. Its speed is zero.

A. I only
B. II only
C. I and II only
D. I and III only

5. Body P increases its velocity from 4.0 m s⁻¹ to 10.0 m s⁻¹ in time $T$ . Body Q increases its velocity from 2.0 m s⁻¹ to 20.0 m s⁻¹ in time $2T$ . What is $\dfrac{\text{acceleration of Q}}{\text{acceleration of P}}$ ? [1]
A. 0.5
B. 1.5
C. 2.0
D. 3.0

6. Define displacement. [1]

7. State what is represented by the gradient of a velocity-time graph and by the area under it. [2]

8. Calculate the average velocity of a ball that moves from $x = -3.0$ m to $x = 5.0$ m in 4.0 s. [2]

9. Explain whether a body with a negative acceleration must be slowing down. [2]

10. A model electric car starts from rest on a level floor and its motor gives it a constant acceleration. A motion sensor records its velocity every 0.50 s. The data below were written for this question; the times can be taken as exact.

t / s

0.50

1.00

1.50

2.00

2.50

v / m s⁻¹ ± 0.02 m s⁻¹

0.37

0.61

0.88

1.12

1.36

(a) Calculate the acceleration of the car from the readings at 0.50 s and 2.50 s, with its absolute uncertainty. [3]
(b) The car was meant to be released at the instant the sensor started recording, $t = 0$ . Determine whether the data are consistent with this. [2]
(c) Suggest what happened, and state the effect on the value of the acceleration found in (a). [2]

11. A puck is pushed up a long, smooth slope of ice and slides back down. Its position up the slope is $x = 6t - t^2$ , with $x$ in m and $t$ in s, shown on the graph.

Position up a slope against time for a puck: a symmetric arch rising from 0 m at 0 s to a peak of 9 m at 3 s and returning to 0 m at 6 s

(a) State the time at which the puck is momentarily at rest, and give your reason. [2]
(b) State the direction of the puck's velocity at $t = 4.0$ s. [1]
(c) The velocity is $v = 6 - 2t$ . Sketch the velocity-time graph from 0 to 6.0 s, with values at the start, the end and where it crosses the time axis. [2]
(d) Deduce the acceleration of the puck, and state whether it changes when the puck turns round. [2]

Try the questions before checking your work.

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