大象传媒

Speed, distance and time

Distance is how far an object moves or how far apart two positions are.

Speed is the of distance - it is the distance travelled per unit time.

The velocity of an object is its speed in a particular direction.

Video: An introduction to the speed equation

In this short video Professor Brian Cox gives an introduction to speed. He defines what is meant by speed and how to calculate speed by applying the 'distance divided by time' equation.

Calculations involving speed, distance and time

The speed of an object can be calculated using the equation:

\(average\) \(speed = \frac{total~distance~travelled}{total~time~taken}\)

\(v = \frac{s}{t}\)

This is when:

  • average speed (v) is measured in metres per second (m / s)

  • total distance travelled (s) is measured in metres (m)

  • total time taken (t) is measured in seconds (s)

Example

A car travels 500 m in 50 s, then 1500 m in 75 s. Calculate its average speed for the whole journey.

First calculate total distance travelled (s):

500 + 1500 = 2000 m

Then calculate total time taken (t):

50 + 75 = 125 s

Then use:

\(v = \frac{s}{t}\)

\(v = 2000 \div 125\)

\(v = 16~m/s\)

Extended syllabus content: Defining acceleration

If you are studying the Extended syllabus, you will also need to define acceleration and use the relevant equation. Click 'show more' for this content:

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Podcast: Displacement, distance and speed

This episode gives an introduction to displacement, distance and speed, including their differences and how to calculate them.

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Distance-time graphs

If an object moves along a straight line, the distance travelled can be represented by a distance-time graph.

Key fact: in a distance-time graph, the gradient of the line is equal to the speed of the object. The greater the gradient (and the steeper the line) the faster the object is moving.

A distance time graph shows distance travelled measured by time.

Example

Calculate the speed of the object represented by the green line in the graph, from 0 to 4 s.

change in distance = (8 - 0) = 8 m

change in time = (4 - 0) = 4 s

\(speed = \frac{distance}{time}\)

\(speed = 8 \div 4\)

\(speed = 2~m/s\)

Question

Calculate the speed of the object represented by the purple line in the graph.

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Distance-time graphs for accelerating objects

If the speed of an object changes, it will be accelerating or . This can be shown as a curved line on a distance-time graph.

A graph to show distance travelled by time. A shows acceleration, B shows constant speed, C shows deceleration, and A shows stationary position. Three dotted lines separate each section.

The table shows what each section of the graph represents:

Section of graphGradientSpeed
AIncreasingIncreasing
BConstantConstant
CDecreasingDecreasing
DZeroStationary (at rest)

If an object is accelerating or decelerating, its speed can be calculated at any particular time by:

  • drawing a to the curve at that time

  • measuring the gradient of the tangent

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Speed-time graphs

Speed-time graphs are useful when describing the movement of an object.

We can use them to determine whether or not the object is moving at any point in time. We can also use them to see what speed the object is travelling at that point in time.

Using data from the graph, we can calculate any , the change in speed and the change in time.

We can also use graphs to calculate distance travelled. The area under a speed-time graph represents the distance travelled.

Extended syllabus content: Speed-time graphs for accelerating objects

If you are studying the Extended syllabus, you will also need to know about speed-time graphs for accelerating objects. Click 'show more' for this content:

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Calculating the area under a speed-time graph

The distance of an object can be calculated from the area under a speed-time graph.

The area under the graph can be calculated by:

  • using geometry (if the lines are straight)

  • counting the squares beneath the line (particularly if the lines are curved)

Example

Calculate the total distance of the object, whose motion is represented by the speed-time graph below.

The y axis shows velocity in metres per second and the x axis time in seconds.  The object increases its velocity from 0 metres per second to 8 metres per second in 4 seconds.

Here, the distance can be found by calculating the total area of the shaded sections below the line.

Find the area of the triangle:

\(\frac{1}{2} \times base \times height\)

\(\frac{1}{2} \times 4 \times 8 = 16\)

The area of the triangle is 16 m2

Find the area of the rectangle:

base 脳 height

(10 - 4) 脳 8 = 48

The area of the rectangle is 48 m2

Add the areas together to find the total distance:

(16 + 48) = 64

Total distance = 64 m

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Free fall

Near the surface of the Earth, any object falling freely will have an acceleration of about 9.8 metres per second squared (m/s2).

Extended syllabus content: The motion of objects falling

If you are studying the Extended syllabus, you will also need to be able to describe the motion of objects falling. Click 'show more' for this content:

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Quiz

Test your knowledge with this quiz on describing motion.

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Teaching resources

Are you a physics teacher looking for more resources? Share these short videos with your students, in which Brian Cox introduces the speed equation and explains the importance of always using common sense and checking that any mathematical answers seem reasonable:

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