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Revision notes for Edexcel GCSE Physics Speed, velocity and acceleration. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Physics (1PH0) specification, so the content matches what's examinable rather than general Physics background.

Speed, velocity and acceleration

What you'll learn

  • How to tell the difference between scalar and vector quantities.
  • How to calculate speed, distance, velocity and acceleration using GCSE equations.
  • How to interpret distance-time and velocity-time graphs.
  • How speeds can be measured in the lab, including using light gates.

Scalars and vectors

Before we describe motion, we need to be clear about what kind of quantity we are using.

Magnitude means the size or amount of something. For example, 5 metres has a magnitude of 5 metres.

Definition

Scalar and vector quantities

A scalar quantity has magnitude but no specific direction. A vector quantity has both magnitude and a specific direction.

Examples you need to know:

  • Distance is scalar; displacement is vector.
  • Speed is scalar; velocity is vector.
  • Mass is scalar; weight is vector.
  • Energy is scalar.
  • Acceleration, force and momentum are vectors.

Distance and displacement

Distance is the total length of the path travelled.
Displacement is the straight-line distance from the starting point to the finishing point, in a stated direction.

Common Mistake

Distance is not displacement

If you walk in a circle and finish where you started, your distance travelled is not zero, but your displacement is zero.

Example

Classifying quantities

A student walks 30 m north, then 30 m south, ending where they started.

  1. The distance is the total path length: 30 m plus 30 m, so the distance is 60 m.
  2. The displacement depends only on start and finish positions. The student finishes where they started, so the displacement is 0 m.
  3. Distance is a scalar because no direction is needed. Displacement is a vector because it must include direction when it is not zero.

Speed and velocity

Speed tells you how quickly distance is covered. It does not include direction, so it is a scalar.

Definition

Velocity

Velocity is speed in a stated direction. For example, “12 m/s east” is a velocity.

For Edexcel 1PH0, you must recall and use these equations:

average speed=distance travelledtime takendistance travelled=average speed×time taken\begin{aligned} \text{average speed} &= \frac{\text{distance travelled}}{\text{time taken}} \\ \text{distance travelled} &= \text{average speed} \times \text{time taken} \end{aligned}average speeddistance travelled​=time takendistance travelled​=average speed×time taken​

Speed is measured in metres per second, written as m/s. Distance is measured in metres, m. Time is measured in seconds, s.

Tip

Unit check

If you calculate speed, your answer should usually have units of m/s. If you calculate distance using speed times time, the seconds cancel, leaving metres.

Example

Calculating distance travelled

A cyclist travels at an average speed of 6 m/s for 45 s. Find the distance travelled.

  1. Use the equation for distance travelled:

    distance=average speed×time\text{distance} = \text{average speed} \times \text{time}distance=average speed×time

  2. Substitute the values with units:

    distance=6 m/s×45 s\text{distance} = 6\ \text{m/s} \times 45\ \text{s}distance=6 m/s×45 s

  3. Calculate and check the unit:

    distance=270 m\text{distance} = 270\ \text{m}distance=270 m

Distance-time graphs

A distance-time graph shows how the distance from a starting point changes over time.

The key idea is that the gradient of a distance-time graph gives the speed.

Distance-time graph showing constant speed, stationary section, higher speed, and gradient as speed

Key Idea

Distance-time graph gradient

On a distance-time graph, a steeper line means a greater speed. A horizontal line means the object is stationary.

For a straight-line section:

speed=change in distancechange in time\text{speed} = \frac{\text{change in distance}}{\text{change in time}}speed=change in timechange in distance​
Example

Finding speed from a distance-time graph

Between 4 s and 10 s, an object’s distance increases from 12 m to 48 m. Find its speed during this section.

  1. Calculate the change in distance:

    48 m−12 m=36 m48\ \text{m} - 12\ \text{m} = 36\ \text{m}48 m−12 m=36 m

  2. Calculate the change in time:

    10 s−4 s=6 s10\ \text{s} - 4\ \text{s} = 6\ \text{s}10 s−4 s=6 s

  3. Use gradient as speed:

    speed=36 m6 s=6 m/s\text{speed} = \frac{36\ \text{m}}{6\ \text{s}} = 6\ \text{m/s}speed=6 s36 m​=6 m/s
Common Mistake

Using the height instead of the gradient

On a distance-time graph, the distance value tells you where the object is. The gradient tells you how fast it is moving.

Acceleration

Acceleration means the rate of change of velocity. In GCSE Physics, this usually means how much the velocity changes each second.

Definition

Acceleration

Acceleration is change in velocity divided by time taken. It is measured in metres per second squared, m/s².

For Edexcel 1PH0, you must recall and use:

a=v−uta = \frac{v-u}{t}a=tv−u​

where:

  • aaa is acceleration in m/s²
  • uuu is initial velocity in m/s
  • vvv is final velocity in m/s
  • ttt is time taken in s
Example

Calculating acceleration

A car speeds up from 5 m/s to 23 m/s in 6 s. Find its acceleration.

  1. Calculate the change in velocity:

    v−u=23 m/s−5 m/s=18 m/sv-u = 23\ \text{m/s} - 5\ \text{m/s} = 18\ \text{m/s}v−u=23 m/s−5 m/s=18 m/s

  2. Divide by the time taken:

    a=18 m/s6 sa = \frac{18\ \text{m/s}}{6\ \text{s}}a=6 s18 m/s​

  3. Calculate the acceleration:

    a=3 m/s2a = 3\ \text{m/s}^2a=3 m/s2
Common Mistake

Acceleration does not always mean speeding up

An object can have negative acceleration if its velocity decreases. Also, because velocity is a vector, changing direction counts as acceleration even if the speed stays the same.

The no-time equation

Sometimes you are given distance but not time. For uniform acceleration in a straight line, you can use:

v2−u2=2×a×xv^2 - u^2 = 2 \times a \times xv2−u2=2×a×x

Here, xxx is the distance travelled in the direction of motion. In the Edexcel spec, this equation is listed as use, not recall and use, so the main skill is knowing how to substitute into it correctly.

Common Mistake

Only for uniform acceleration

Use v2−u2=2axv^2 - u^2 = 2axv2−u2=2ax only when the acceleration is constant and the motion is in a straight line.

Example

Finding final velocity without time

A trolley starts from rest and accelerates uniformly at 2 m/s² over a distance of 9 m. Find its final velocity.

  1. Since the trolley starts from rest, the initial velocity is:

    u=0 m/su = 0\ \text{m/s}u=0 m/s

  2. Substitute into the equation:

    v2−02=2×2 m/s2×9 mv^2 - 0^2 = 2 \times 2\ \text{m/s}^2 \times 9\ \text{m}v2−02=2×2 m/s2×9 m

  3. Calculate v2v^2v2, then take the square root:

    v2=36 m2/s2v^2 = 36\ \text{m}^2\text{/s}^2v2=36 m2/s2 v=6 m/sv = 6\ \text{m/s}v=6 m/s

Velocity-time graphs

A velocity-time graph shows how velocity changes over time.

The gradient of a velocity-time graph gives acceleration. The area between the graph line and the time axis gives the distance travelled, for the uniform acceleration sections you meet in this topic.

Velocity-time graph showing gradient as acceleration and shaded area as distance travelled

Key Idea

Velocity-time graphs

On a velocity-time graph, gradient = acceleration and area under the graph = distance travelled.

Example

Using a velocity-time graph

An object’s velocity increases uniformly from 4 m/s to 12 m/s in 5 s. Find its acceleration and distance travelled.

  1. Use the gradient to find acceleration:

    a=12 m/s−4 m/s5 s=1.6 m/s2a = \frac{12\ \text{m/s} - 4\ \text{m/s}}{5\ \text{s}} = 1.6\ \text{m/s}^2a=5 s12 m/s−4 m/s​=1.6 m/s2

  2. Use the area under the graph to find distance. The shape is a trapezium with parallel sides 4 and 12, and width 5:

    distance=4+122×5\text{distance} = \frac{4 + 12}{2} \times 5distance=24+12​×5

  3. Calculate the distance:

    distance=40 m\text{distance} = 40\ \text{m}distance=40 m
Common Mistake

Mixing up the two graph rules

Distance-time graph: gradient gives speed. Velocity-time graph: gradient gives acceleration, and area gives distance.

Measuring speed in the lab

You should be able to describe methods for finding the speed of an object.

Stopwatch and measured distance

You can measure a distance with a metre ruler or tape measure, then time how long an object takes to travel that distance.

This gives average speed:

average speed=distancetime\text{average speed} = \frac{\text{distance}}{\text{time}}average speed=timedistance​

To improve the result, use a longer distance, repeat the measurement, and calculate a mean.

Light gates

A light gate uses a beam of light connected to a timer. When an object or card blocks the beam, the timer records how long the beam is interrupted.

If you know the length of the card:

speed=card lengthtime blocking the light gate\text{speed} = \frac{\text{card length}}{\text{time blocking the light gate}}speed=time blocking the light gatecard length​
Example

Calculating speed using a light gate

A card of length 0.080 m passes through a light gate. It blocks the beam for 0.020 s. Find the speed of the card.

  1. Use the card length as the distance that passes through the beam:

    distance=0.080 m\text{distance} = 0.080\ \text{m}distance=0.080 m

  2. Substitute into the speed equation:

    speed=0.080 m0.020 s\text{speed} = \frac{0.080\ \text{m}}{0.020\ \text{s}}speed=0.020 s0.080 m​

  3. Calculate the speed:

    speed=4.0 m/s\text{speed} = 4.0\ \text{m/s}speed=4.0 m/s
Tip

Why light gates are useful

Light gates reduce human reaction-time error because the timer starts and stops electronically.

Other useful methods include motion sensors, video analysis, and ticker timers.

Typical speeds and free fall

You should know some rough everyday speeds. These are approximate, but useful for checking whether an answer is sensible:

  • Walking: about 1 to 2 m/s
  • Running: about 3 to 6 m/s
  • Cycling: about 5 to 8 m/s
  • Car in town: about 13 m/s
  • Motorway car: about 30 m/s
  • Sound in air: about 330 m/s
  • Wind: a few m/s for a breeze, much higher in storms

Near the Earth’s surface, the acceleration due to gravity is:

g=10 m/s2g = 10\ \text{m/s}^2g=10 m/s2

This means that in free fall, ignoring air resistance, an object’s downward velocity increases by about 10 m/s every second.

Example

Estimating an everyday acceleration

A bus increases its velocity from 0 m/s to 15 m/s in 6 s. Estimate its acceleration.

  1. Calculate the change in velocity:

    15 m/s−0 m/s=15 m/s15\ \text{m/s} - 0\ \text{m/s} = 15\ \text{m/s}15 m/s−0 m/s=15 m/s

  2. Divide by the time taken:

    a=15 m/s6 s=2.5 m/s2a = \frac{15\ \text{m/s}}{6\ \text{s}} = 2.5\ \text{m/s}^2a=6 s15 m/s​=2.5 m/s2

  3. Compare with ggg:

    2.5 m/s22.5\ \text{m/s}^22.5 m/s2

    is much smaller than 10 m/s210\ \text{m/s}^210 m/s2, so it is a realistic acceleration for a vehicle.

Exam technique

In the exam

  1. Decide whether the quantity is scalar or vector: if direction matters, it is a vector.
  2. For distance-time graphs, use the gradient to find speed.
  3. For velocity-time graphs, use the gradient for acceleration and the area for distance.
  4. Always convert units before substituting, especially kilometres to metres and minutes to seconds.
  5. Check whether acceleration is uniform before using v2−u2=2axv^2 - u^2 = 2axv2−u2=2ax.
Self review

Check yourself

  • What is the difference between speed and velocity?
  • How do you find speed from a distance-time graph?
  • What does the area under a velocity-time graph represent?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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