3.1.1 - Kinematics and motion graphs | OCR A-Level Physics

Motion is easier to analyse when you can move between words, equations and graphs. In this lesson you will define the main kinematics quantities, interpret displacement-time and velocity-time graphs, and use gradients and areas to extract physical information. These skills are also practical skills: a motion sensor or data-logger only becomes useful when you know what the graph means.

Motion Quantities

Kinematics describes motion without first asking what force caused it. For this lesson, the most important quantities are displacement, speed, velocity and acceleration. They sound familiar, but each has a precise meaning.

Displacement

Displacement is the change in position from a chosen starting point to a final position, in a specified direction. It is a vector quantity, so direction matters.

Distance travelled and displacement are not the same. If a trolley moves $3.0\,\mathrm{m}$ east and then $3.0\,\mathrm{m}$ west back to its start, the total distance travelled is $6.0\,\mathrm{m}$ , but the displacement from the start is $0\,\mathrm{m}$ .

Speed and velocity make the same scalar-vector contrast.

Speed

Speed is the rate of change of distance. It is a scalar quantity and is never negative.

Speed tells you how fast distance is being covered, but it does not tell you which way the object is moving.

Velocity

Velocity is the rate of change of displacement. It is a vector quantity, so its sign or direction depends on the chosen positive direction.

Average speed uses total distance travelled:

Average Speed

\text{average speed}=\frac{\text{total distance travelled}}{\text{time taken}}

Average velocity uses displacement:

Average Velocity

\text{average velocity}=\frac{\text{displacement}}{\text{time taken}}

The units are important. Distance and displacement are measured in metres, $\mathrm{m}$ . Speed and velocity are measured in metres per second, $\mathrm{m} \mathrm{s}^{-1}$ .

Worked example: average speed and average velocity

A student walks $12\,\mathrm{m}$ east in $6.0\,\mathrm{s}$ , then $4.0\,\mathrm{m}$ west in the next $4.0\,\mathrm{s}$ .

Total distance travelled:

12+4.0=16\ \text{m}

Total displacement, taking east as positive:

+12-4.0=+8.0\ \text{m}

Total time:

6.0+4.0=10.0\ \text{s}

Average speed:

\frac{16}{10.0}=1.6\ \text{m s}^{-1}

Average velocity:

\frac{+8.0}{10.0}=+0.80\ \text{m s}^{-1}

The plus sign means the average velocity is east, using the sign convention chosen in the question.

Acceleration describes how velocity changes.

Acceleration

Acceleration is the rate of change of velocity. It is a vector quantity with unit $\mathrm{m} \mathrm{s}^{-2}$ .

For a time interval where the change in velocity is known:

Acceleration

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

Here, $a$ is acceleration in $\mathrm{m} \mathrm{s}^{-2}$ , $\Delta v$ is the change in velocity in $\mathrm{m} \mathrm{s}^{-1}$ , and $\Delta t$ is the time interval in $\mathrm{s}$ .

Worked example: acceleration from a change in velocity

A model car's velocity changes from $+2.0\,\mathrm{m}\,\mathrm{s}^{-1}$ to $+8.0\,\mathrm{m}\,\mathrm{s}^{-1}$ in $3.0\,\mathrm{s}$ .

\Delta v=+8.0-(+2.0)=+6.0\ \text{m s}^{-1}

a=\frac{+6.0}{3.0}=+2.0\ \text{m s}^{-2}

The positive acceleration means the velocity is becoming more positive. It does not simply mean "moving to the right" unless the positive direction has been defined that way.

An order-of-magnitude check uses powers of ten before detailed arithmetic. A journey of about $10^3\,\mathrm{m}$ lasting $10^2\,\mathrm{s}$ has an average speed of order $10\,\mathrm{m\,s^{-1}}$ . An answer of $10^4\,\mathrm{m\,s^{-1}}$ would therefore suggest a unit conversion or power-of-ten error. As a rough comparison, walking speeds are of order $1\,\mathrm{m\,s^{-1}}$ , road-vehicle speeds of order $10\,\mathrm{m\,s^{-1}}$ , and aircraft speeds of order $10^2\,\mathrm{m\,s^{-1}}$ . These are scales, not exact limits.

Instantaneous And Average Rates

A rate of change compares how quickly one quantity changes as another quantity changes. In kinematics, the second quantity is usually time.

An average rate is calculated over a whole interval. An instantaneous rate is the rate at one particular instant. This distinction is one of the most important ideas in this lesson.

Instantaneous Speed

Instantaneous speed is the speed of an object at a particular instant. It is the magnitude of the instantaneous velocity.

For example, the reading on a car speedometer is an instantaneous speed. It does not tell you the average speed for the whole journey.

Worked example: average rate from two readings

A trolley has displacement $0.50\,\mathrm{m}$ at $t = 1.0\,\mathrm{s}$ and displacement $2.90\,\mathrm{m}$ at $t = 4.0\,\mathrm{s}$ .

The average velocity between these times is:

\frac{\Delta s}{\Delta t}=\frac{2.90-0.50}{4.0-1.0}=\frac{2.40}{3.0}=0.80\ \text{m s}^{-1}

This is an average over the $3.0\,\mathrm{s}$ interval. The trolley may have been moving slower than this at the start and faster than this at the end.

In graph language, an average rate is the gradient of a chord joining two points. An instantaneous rate is the gradient of a tangent at one point.

[DIAGRAM: asset_name: 3.1.1 - Kinematics and motion graphs - diagram 1; asset_slug: 3_01_1_kinematics_graphs_and_rates__diagram_01; recommended_method: drawn_physics; caption: A chord gives average velocity and a tangent gives instantaneous velocity on a displacement-time graph.; description: A curved displacement-time graph with an endpoint chord for average velocity and a tangent for instantaneous velocity.]

The units reveal the physical meaning of the gradient:

\frac{\text{vertical-axis unit}}{\text{horizontal-axis unit}}=\frac{\text{m}}{\text{s}}=\text{m s}^{-1}

So the gradient of a displacement-time graph has the unit of velocity.

Displacement-Time Graphs

On a displacement-time graph, the vertical axis is displacement and the horizontal axis is time. The graph does not show the path taken through space. It shows how displacement from a chosen origin changes with time.

The gradient gives velocity:

Velocity From A Displacement-Time Graph

v=\frac{\Delta s}{\Delta t}

For a straight section, the velocity is constant because the gradient is constant. For a curve, the velocity is changing because the gradient changes.

Useful graph meanings:

Displacement-time feature	Motion meaning
horizontal line	zero velocity; displacement is constant
straight line with positive gradient	constant positive velocity
straight line with negative gradient	constant negative velocity
steeper gradient	larger magnitude of velocity
curve getting steeper	speed increasing in that direction
curve becoming flatter	speed decreasing in that direction

Worked example: velocity from a straight displacement-time section

A graph shows displacement increasing from $1.0\,\mathrm{m}$ to $7.0\,\mathrm{m}$ between $t = 2.0\,\mathrm{s}$ and $t = 5.0\,\mathrm{s}$ .

\Delta s=7.0-1.0=6.0\ \text{m}

\Delta t=5.0-2.0=3.0\ \text{s}

v=\frac{6.0}{3.0}=2.0\ \text{m s}^{-1}

Because the section is straight, this is the velocity throughout that section.

Worked example: estimating instantaneous velocity from a tangent

A curved displacement-time graph has a tangent drawn at $t = 3.0\,\mathrm{s}$ . Two clear points on the tangent are $(2.0\,\mathrm{s}, 1.6\,\mathrm{m})$ and $(4.0\,\mathrm{s}, 5.0\,\mathrm{m})$ .

Use the tangent, not nearby raw points from the curve:

v\approx\frac{5.0-1.6}{4.0-2.0}=\frac{3.4}{2.0}=1.7\ \text{m s}^{-1}

The word "approximately" matters because drawing and reading a tangent from a curve involves judgement.

A common error is to say that a curved displacement-time graph means a curved path. It does not. A ball moving in a straight line can have a curved displacement-time graph if its velocity changes.

Velocity-Time Graphs

On a velocity-time graph, the vertical axis is velocity and the horizontal axis is time. This graph gives two different pieces of physical information.

The gradient gives acceleration:

Acceleration From A Velocity-Time Graph

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

The area under the graph gives displacement:

Displacement From A Velocity-Time Graph

\text{displacement}=\text{area under the velocity-time graph}

The unit check is the quickest way to remember the area result:

\text{area unit}=(\text{m s}^{-1})\times\text{s}=\text{m}

The area has the unit of displacement.

[DIAGRAM: asset_name: 3.1.1 - Kinematics and motion graphs - diagram 2; asset_slug: 3_01_1_kinematics_graphs_and_rates__diagram_02; recommended_method: drawn_physics; caption: Gradient gives acceleration and signed area gives displacement on a velocity-time graph.; description: A velocity-time graph with a gradient triangle for acceleration and a shaded area for displacement.]

Worked example: gradient and area from a velocity-time graph

A toy car has the following velocity-time graph:

from $0\,\mathrm{s}$ to $2.0\,\mathrm{s}$ , velocity increases uniformly from $0$ to $6.0\,\mathrm{m}\,\mathrm{s}^{-1}$
from $2.0\,\mathrm{s}$ to $5.0\,\mathrm{s}$ , velocity is constant at $6.0\,\mathrm{m}\,\mathrm{s}^{-1}$

Acceleration in the first section:

a=\frac{6.0-0}{2.0-0}=3.0\ \text{m s}^{-2}

Displacement in the first section is the triangular area:

\frac{1}{2}\times 2.0\times 6.0=6.0\ \text{m}

Displacement in the second section is the rectangular area:

3.0\times 6.0=18\ \text{m}

Total displacement:

6.0+18=24\ \text{m}

This method works because each small area has the form velocity times time.

If the graph goes below the time axis, the velocity is negative. The area below the axis represents negative displacement. If a question asks for total distance travelled, you must add the magnitudes of the separate areas. If it asks for displacement, keep the signs.

Curves And Data-Loggers

Real motion data is often curved because velocity or acceleration changes continuously. A data-logger, light gate system, video analysis program or motion sensor can produce displacement-time or velocity-time data very quickly. The physics skill is still the same: interpret the graph.

For a curved displacement-time graph:

draw a tangent at the time of interest;
use two well-spaced points on the tangent;
calculate the tangent gradient to estimate instantaneous velocity.

For a curved velocity-time graph:

the tangent gradient estimates instantaneous acceleration;
the area under the curve estimates displacement;
a finer grid, more strips or more frequent data samples usually improves the estimate.

[DIAGRAM: asset_name: 3.1.1 - Kinematics and motion graphs - diagram 3; asset_slug: 3_01_1_kinematics_graphs_and_rates__diagram_03; recommended_method: drawn_physics; caption: Four trapezia estimate a displacement of 17 m from the velocity-time curve.; description: Four trapezia approximate the area below a velocity-time curve, giving a displacement of 17 m.]

Worked example: estimating area under a non-linear velocity-time graph

A motion sensor gives these velocity values:

$t /\mathrm{s}$	$0$	$1$	$2$	$3$	$4$
$v /\mathrm{m} s^{-1}$	$0.0$	$3.0$	$5.0$	$6.0$	$6.0$

Estimate the displacement from $0\,\mathrm{s}$ to $4\,\mathrm{s}$ using trapezia of width $1\,\mathrm{s}$ .

First strip, from $0\,\mathrm{s}$ to $1\,\mathrm{s}$ :

\frac{1}{2}(0.0+3.0)\times 1=1.5\ \text{m}

Second strip:

\frac{1}{2}(3.0+5.0)\times 1=4.0\ \text{m}

Third strip:

\frac{1}{2}(5.0+6.0)\times 1=5.5\ \text{m}

Fourth strip:

\frac{1}{2}(6.0+6.0)\times 1=6.0\ \text{m}

Estimated displacement:

1.5+4.0+5.5+6.0=17.0\ \text{m}

This is an estimate of the area under the non-linear graph. A data-logger with more frequent samples would let you use narrower strips and usually reduce the area-estimation uncertainty.

Graph presentation still matters in practical work. Use a conventional scale increasing left to right, label each axis with quantity and unit, and show gradient working clearly. If you use a gradient triangle, make it large enough to read accurately and use points on the line or tangent.

Speed, velocity and acceleration graphs

Speed–time and acceleration–time graphs complete the description of a journey. Speed is the magnitude of velocity, so a speed–time graph cannot go below zero. To turn a velocity–time graph into a speed–time graph, reflect each negative-velocity section above the time axis. Its area gives total distance, while the signed velocity–time area gives displacement.

An acceleration–time graph shows how rapidly velocity changes. A horizontal line above zero means constant positive acceleration; a line on zero means constant velocity, which need not be zero. The signed area gives the change in velocity because acceleration multiplied by time has units of velocity:

\Delta v=\text{signed area under an acceleration–time graph}.

For example, acceleration $+2.0\,\mathrm{m\,s^{-2}}$ for $3.0\,\mathrm{s}$ changes velocity by $+6.0\,\mathrm{m\,s^{-1}}$ . If the initial velocity is $-8.0\,\mathrm{m\,s^{-1}}$ , the final velocity is $-2.0\,\mathrm{m\,s^{-1}}$ : positive acceleration has reduced the speed. Acceleration and velocity in the same direction increase speed; opposite directions decrease it until any reversal.

[DIAGRAM: asset_name: Kinematics graph comparison; asset_slug: mechanics_speed_acceleration; recommended_method: drawn_physics; caption: Velocity, speed and acceleration graphs describe the same object reversing direction.; description: Signed velocity, nonnegative speed and acceleration are plotted against the same time for a reversing object.]

A dimensional check helps choose the graph operation. Displacement divided by time is velocity; velocity divided by time is acceleration. Area under displacement–time has units $\mathrm{m\,s}$ , so it is not velocity.

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3.1.1 - Kinematics and motion graphs