Edexcel IGCSE Physics 1(b) Movement and Position — Notes
Written by an examiner for Pearson Edexcel International GCSE Physics (4PH1), Science Double Award (4SD0) and Science Single Award (4SS0), linear and modular. This page also covers the 1(a) units for Topic 1.
Which parts do I need?
Single Award candidates can skip every section tagged Double · Physics. On this page that is only the equation \(v^2=u^2+2as\) and the Topic 1 units for momentum and moments.
What this page covers
| Spec | You need to be able to… | Route |
|---|---|---|
| 1.1 | use kg, m, m/s, m/s², N, s and N/kg | Single · Double · Physics |
| 1.2P | use N m and kg m/s | Physics only · 2P |
| 1.3 | plot distance–time graphs and explain what they show | Single · Double · Physics |
| 1.4 | recall and use average speed = distance moved ÷ time taken | Single · Double · Physics |
| 1.5 | investigate the motion of everyday objects (practical) | Single · Double · Physics |
| 1.6 | recall and use acceleration = change in velocity ÷ time taken | Single · Double · Physics |
| 1.7 | plot velocity–time graphs and explain what they show | Single · Double · Physics |
| 1.8 | find acceleration from the gradient of a velocity–time graph | Single · Double · Physics |
| 1.9 | find the distance travelled from the area under a velocity–time graph | Single · Double · Physics |
| 1.10 | use \(v^2=u^2+2as\) | Double · Physics |
Modular candidates: these are statements 1.1–1.10 of 4XPH1 Unit 1 (4WPH1) and 4XSD1 Physics Unit 5 (4WSD5).
In one minute
- Average speed = distance moved ÷ time taken, in m/s.
- On a distance–time graph, the gradient is the speed. A horizontal line means the object is stationary.
- Acceleration = change in velocity ÷ time taken, \(a=\dfrac{v-u}{t}\), in m/s². A negative value means the object is slowing down (decelerating).
- On a velocity–time graph, the gradient is the acceleration and the area under the graph is the distance travelled.
- For uniform acceleration, \(v^2=u^2+2as\) links speeds and distance without needing the time. Double · Physics
Which units do I need for Topic 1? Single · Double · Physics
Every answer to a calculation needs a unit. These are the units Topic 1 uses; learn the symbol and what each one measures.
| Quantity | Unit | Symbol | Route |
|---|---|---|---|
| mass | kilogram | kg | All |
| distance, length | metre | m | All |
| time | second | s | All |
| speed, velocity | metre/second | m/s | All |
| acceleration | metre/second² | m/s² | All |
| force, weight | newton | N | All |
| gravitational field strength | newton/kilogram | N/kg | All |
| moment of a force | newton metre | N m | Physics only · 2P |
| momentum | kilogram metre/second | kg m/s | Physics only · 2P |
- Convert before you substitute. Change km to m (× 1000), minutes to s (× 60) and hours to s (× 3600).
- To change a speed from km/h to m/s, divide by 3.6. For example, 54 km/h = 54 × 1000 ÷ 3600 = 15 m/s.
- Moments (N m) and momentum (kg m/s) are taught in 1(c); they appear here only because the specification lists all Topic 1 units together.
What does a distance–time graph show? Single · Double · Physics
A distance–time graph plots the distance an object has moved (vertical axis) against time (horizontal axis). Its shape tells you how the object is moving.
| Shape of the line | What the object is doing |
|---|---|
| horizontal | stationary (speed = 0) |
| straight and sloping | moving at a constant speed |
| steeper straight line | moving at a greater constant speed |
| curving upwards (getting steeper) | speeding up — accelerating |
| curving and levelling off (getting less steep) | slowing down — decelerating |
Finding the speed from the graph
- For a straight section, pick two points far apart and calculate \(\text{speed}=\dfrac{\text{change in distance}}{\text{change in time}}\).
- Draw the triangle on the graph so the examiner can see which values you read.
- The average speed for a whole journey is the total distance divided by the total time, including any time spent stationary.
Plotting a graph that earns full marks
- Put time on the horizontal axis. Label both axes with the quantity and the unit, for example “distance / m”.
- Choose a linear scale that uses more than half of the grid in each direction. Avoid awkward scales such as 3 or 7 units per square.
- Plot every point to within half a small square, using a small cross or dot.
- Draw a single straight line or smooth curve of best fit. Do not join the points dot to dot.
How do you calculate average speed? Single · Double · Physics
\[ \text{average speed}=\frac{\text{distance moved}}{\text{time taken}} \]- Speed in m/s, distance in m, time in s.
- “Average” matters: during a journey the speed changes, so this gives the constant speed that would cover the same distance in the same time.
- Rearranged: distance = average speed × time; time = distance ÷ average speed.
A cyclist rides 4.2 km in 15 minutes. Calculate her average speed in m/s.
Convert first: distance = 4.2 × 1000 = 4200 m; time = 15 × 60 = 900 s.
average speed = 4200 ÷ 900 = 4.67 = 4.7 m/s
The graph shows a walker's journey in three stages, A, B and C.
(a) Describe the motion in stage B. (b) Calculate the speed in stage A and in stage C. (c) Calculate the average speed for the whole journey.
(a) The line is horizontal, so the distance does not change: the walker is stationary for 30 s.
(b) Stage A: speed = 60 ÷ 20 = 3.0 m/s. Stage C: speed = (180 − 60) ÷ (80 − 50) = 120 ÷ 30 = 4.0 m/s.
Stage C is steeper than stage A, which matches the greater speed.
(c) average speed = total distance ÷ total time = 180 ÷ 80 = 2.25 m/s
The stationary time in stage B counts in the total time, which is why the average (2.25 m/s) is lower than the speed in either moving stage.
Practical: how do you investigate the motion of everyday objects? Single · Double · Physics
The specification asks you to investigate the motion of objects such as toy cars or tennis balls. Questions on this practical ask you to describe a method, name variables, process results and suggest improvements.
Investigating how the speed of a toy car depends on the height of the ramp
- Aim: find how the average speed of a toy car between two points depends on the height from which it is released.
- Apparatus: ramp, toy car, metre rule, two light gates and a data logger (or a stopwatch), blocks to set the ramp height.
- Independent variable: release height of the car (raise the ramp with blocks; measure the height with the metre rule).
- Dependent variable: average speed between the two gates.
- Control variables: the same car, the same release point, the same distance between the gates, the same ramp surface.
Method
- Measure the distance between the two light gates with the metre rule.
- Release the car from rest at a marked line — do not push it.
- Record the time taken to travel between the gates.
- Repeat three times for each height and calculate the mean time.
- Calculate average speed = distance between gates ÷ mean time. Repeat for at least five different heights.
- Plot a graph of average speed (vertical axis) against height (horizontal axis) and draw a line or curve of best fit.
Errors and improvements
- Timing with a stopwatch is affected by human reaction time (about 0.2 s). Light gates remove this, and so does filming the car next to a ruler and stepping through the video frame by frame.
- Use a longer distance between the measuring points, so the time measured is large compared with the reaction time.
- Repeat readings and calculate a mean to reduce the effect of random errors; ignore anomalous results when calculating the mean.
- Safety: place a soft barrier at the bottom of the ramp to stop the car, and keep the ramp stable so it cannot fall.
For a tennis ball, the same ideas apply: drop it from a measured height next to a metre rule, film it, and read its position at equal time intervals from the video to plot a distance–time graph.
What is acceleration and how do you calculate it? Single · Double · Physics
- \(a\) = acceleration (m/s²), \(v\) = final velocity (m/s), \(u\) = initial velocity (m/s), \(t\) = time taken (s).
- An acceleration of 3 m/s² means the velocity increases by 3 m/s every second.
- If the object is slowing down, \(v\) is less than \(u\) and \(a\) is negative. This is called a deceleration.
- Velocity is speed in a stated direction, so a change of direction is also a change of velocity.
(a) A car speeds up from 4.0 m/s to 22 m/s in 6.0 s. Calculate its acceleration.
(b) A bus travelling at 15 m/s stops in 5.0 s. Calculate its acceleration.
(a) \(a=\dfrac{v-u}{t}=\dfrac{22-4.0}{6.0}=\) 3.0 m/s²
(b) \(a=\dfrac{0-15}{5.0}=\) −3.0 m/s², a deceleration of 3.0 m/s².
What does a velocity–time graph show? Single · Double · Physics
A velocity–time graph plots velocity (vertical axis) against time. It carries two pieces of information — one from its gradient and one from its area.
| Shape of the line | What the object is doing |
|---|---|
| along the time axis (v = 0) | at rest |
| horizontal, above the axis | constant velocity (zero acceleration) |
| straight, sloping upwards | uniform (constant) acceleration |
| straight, sloping downwards | uniform deceleration |
| curved | acceleration that is changing (non-uniform) |
- Gradient: \(\text{acceleration}=\dfrac{\text{change in velocity}}{\text{change in time}}\), read from a large triangle on a straight section.
- Area: split the region under the graph into rectangles (base × height) and triangles (½ × base × height), then add them.
- Do not confuse the two graphs: a horizontal line on a distance–time graph means stationary; on a velocity–time graph it means constant velocity.
The graph shows the velocity of a train as it travels between two stations.
(a) Determine the acceleration in stage A and in stage C. (b) Determine the distance between the two stations. (c) Calculate the average speed of the train for the journey.
(a) Stage A: \(a=\dfrac{12-0}{8.0}=\) 1.5 m/s². Stage C: \(a=\dfrac{0-12}{34-28}=\) −2.0 m/s² (the train is decelerating).
(b) Distance = area under the graph:
stage A (triangle) = ½ × 8.0 × 12 = 48 m
stage B (rectangle) = 20 × 12 = 240 m
stage C (triangle) = ½ × 6.0 × 12 = 36 m
total = 48 + 240 + 36 = 324 m
(c) average speed = 324 ÷ 34 = 9.53 = 9.5 m/s
Motion graphs simulation: move an object and watch its distance–time and velocity–time graphs build up together — gradients and areas live.
How do you use \(v^2=u^2+2as\)? Double · Physics
Not required for Single Award (4SS0).
\[ (\text{final speed})^2=(\text{initial speed})^2+(2\times\text{acceleration}\times\text{distance moved})\qquad v^2=u^2+2as \]- Use it only when the acceleration is uniform, and when the question gives (or asks for) speeds and a distance but no time.
- \(v\) and \(u\) in m/s, \(a\) in m/s², \(s\) in m.
- For an object slowing down, \(a\) is negative. When it stops, \(v=0\).
- The equation is on the Equation List, but you must still choose it and rearrange it yourself.
(a) A cyclist moving at 3.0 m/s accelerates uniformly at 0.80 m/s² for 40 m. Calculate her final speed.
(b) A car travelling at 20 m/s brakes with a uniform deceleration of 5.0 m/s². Calculate the distance it travels before it stops.
(a) \(v^2=u^2+2as=3.0^2+2\times0.80\times40=9.0+64=73\)
\(v=\sqrt{73}=8.54=\) 8.5 m/s
(b) \(v=0\), \(u=20\ \text{m/s}\), \(a=-5.0\ \text{m/s}^2\):
\(0=20^2+2\times(-5.0)\times s\Rightarrow s=\dfrac{400}{10}=\) 40 m
Common error in (a): forgetting the square root and giving 73 m/s.
Examiner tips for 1(b)
Many calculation questions begin by asking you to state the relationship, for one mark. Write it in words (average speed = distance moved ÷ time taken) or in standard symbols. Units in place of quantities (m/s = m ÷ s) do not earn the mark.
Write the numbers into the equation before giving the answer. If your final answer is wrong, a clear substitution can still earn a mark; a bare number earns nothing.
Before you start, ask which quantity the question wants. Acceleration comes from the gradient of a velocity–time graph; distance comes from the area. Reading a single value off the graph gives neither.
When asked to describe the motion shown by a graph, give each stage with its values: “accelerates uniformly from rest to 12 m/s in 8 s”, not just “speeds up”. Use the data on the graph.
Check that all values are in m, s and m/s before substituting. A distance in km or a time in minutes is the most common reason for a power-of-ten error.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A horizontal line on any motion graph means the object has stopped. | It depends on the graph. | Distance–time: stationary. Velocity–time: constant velocity. |
| A sloping line on a velocity–time graph means constant speed. | The velocity is changing along a sloping line. | A sloping straight line means uniform acceleration or deceleration. |
| Average speed is the mean of the start and end speeds. | That is only true for uniform acceleration. | Average speed = total distance ÷ total time. |
| Stopping time is left out of the average speed. | Time spent stationary is still part of the journey time. | Divide by the total time, including stops. |
| Deceleration is a positive number in \(a=\frac{v-u}{t}\). | If \(v<u\) the subtraction gives a negative value. | Slowing down gives a negative acceleration. |
Key equations
| Equation | Use it to find | Route | On the Equation List? |
|---|---|---|---|
| average speed = distance moved ÷ time taken | speed from a distance and a time | All | yes — all papers |
| \(a=\dfrac{v-u}{t}\) | acceleration from a change in velocity | All | yes — all papers |
| gradient of distance–time graph | speed | All | — (graph skill) |
| gradient of velocity–time graph | acceleration | All | — (graph skill) |
| area under velocity–time graph | distance travelled | All | — (graph skill) |
| \(v^2=u^2+2as\) | a speed or a distance when the time is not given | Double · Physics | yes — all papers |
Pearson includes the Equation List with the question papers, but the first mark in many questions is for writing the relationship yourself — learn them anyway.
Check your understanding
1. A distance–time graph is a horizontal line for 10 s. What is the object doing? All
It is stationary — the distance does not change, so the speed is zero.
2. A runner covers 1.2 km in 4.0 minutes. Calculate her average speed. All
1200 m ÷ 240 s = 5.0 m/s. (1 mark for converting both units, 1 for substitution, 1 for the answer.)
3. A sprinter goes from rest to 9.0 m/s in 1.8 s. Calculate the acceleration. All
\(a=\dfrac{9.0-0}{1.8}=\) 5.0 m/s²
4. An object accelerates uniformly from rest to 8.0 m/s in 4.0 s. Use the area under the velocity–time graph to find the distance it travels. All
The graph is a triangle: ½ × 4.0 × 8.0 = 16 m.
5. A velocity–time graph and a distance–time graph each show a straight line sloping upwards. Explain how the motion they describe is different. All
The distance–time graph shows a constant speed (constant gradient = constant speed). The velocity–time graph shows an increasing velocity — the object is accelerating uniformly (constant gradient = constant acceleration).
6. An aircraft starts from rest and reaches 72 m/s after travelling 1200 m along the runway. Calculate its acceleration, assuming it is uniform. Double · Physics
\(v^2=u^2+2as\Rightarrow 72^2=0+2\times a\times1200\Rightarrow a=\dfrac{5184}{2400}=2.16=\) 2.2 m/s²