At a glance
- Cambridge section
- 0625 topic 1.2 Motion (Core 1 to 8, Supplement 9 to 13)
- Edexcel section
- 4PH1 statements 1.3 to 1.10, plus 1.21 on terminal velocity
- Speed
- v = s / t (m/s)
- Acceleration
- a = (v - u) / t (m/s^2)
- Edexcel extra
- v^2 = u^2 + 2as
- Free fall
- g is about 9.8 m/s^2 near the Earth's surface
Key ideas, board by board
Speed is the distance travelled per unit time. Velocity is speed in a given direction, so it is a vector: two cars moving at 20 m/s in opposite directions have the same speed but different velocities. Average speed is total distance travelled divided by total time taken, which is not the same as the average of two speeds when the times differ.
Cambridge 0625 Core (1.2.1 to 1.2.8) asks you to define speed and velocity, sketch, plot and interpret distance-time and speed-time graphs, say from a graph when an object is at rest, moving at constant speed, accelerating or decelerating, find speed from the gradient of a distance-time graph, find distance from the area under a speed-time graph, and state that the acceleration of free fall near the Earth is about 9.8 m/s^2. Supplement (1.2.9 to 1.2.13) adds the definition and equation for acceleration, gradients of speed-time graphs, changing acceleration, deceleration as negative acceleration, and falling objects with and without air resistance, including terminal velocity.
Edexcel 4PH1 is not tiered, so every student needs the whole topic: distance-time graphs (1.3), average speed (1.4), acceleration a = (v - u) / t (1.6), velocity-time graphs with gradient and area (1.7 to 1.9), and the equation (final speed)^2 = (initial speed)^2 + 2 x acceleration x distance (1.10). The practical in 1.5 is investigating the motion of everyday objects such as toy cars or tennis balls. Terminal velocity sits in the forces section (1.21). Appendix 7 of the specification lists speed and acceleration as relationships students must recall; v^2 = u^2 + 2as is not on that recall list.
Reading the two graphs
Check the label on the vertical axis before you say anything about a graph. Half the errors in this topic come from treating a speed-time graph like a distance-time graph.
| What you see | Distance-time graph means | Speed-time graph means |
|---|---|---|
| Horizontal line | Stationary (at rest) | Constant speed |
| Straight sloping line upwards | Constant speed; gradient = speed | Constant acceleration; gradient = acceleration |
| Line curving upwards, getting steeper | Accelerating | Acceleration increasing (Cambridge Supplement: changing acceleration) |
| Straight line sloping downwards | Moving back towards the start | Constant deceleration |
| Area under the line | No meaning at IGCSE | Distance travelled |
Worked example 1: a speed-time graph
Question: a car starts from rest and reaches 12 m/s after 4 s. It travels at 12 m/s for 6 s, then slows steadily to a stop in 3 s. Calculate (a) the acceleration in the first 4 s, (b) the total distance travelled, (c) the average speed for the whole journey.
(a) a = (v - u) / t = (12 - 0) / 4 = 3 m/s^2.
(b) Distance is the area under the graph, split into three shapes. First triangle: 1/2 x 4 x 12 = 24 m. Rectangle: 6 x 12 = 72 m. Last triangle: 1/2 x 3 x 12 = 18 m. Total = 24 + 72 + 18 = 114 m.
(c) Average speed = total distance / total time = 114 / 13 = 8.8 m/s (2 significant figures). The deceleration in the last part is (0 - 12) / 3 = -4 m/s^2; the minus sign shows the car is slowing down, which Cambridge Supplement 1.2.12 asks you to use in calculations.
Worked example 2: stopping distance with v^2 = u^2 + 2as (Edexcel)
Question: a car travelling at 20 m/s brakes with a constant deceleration of 4 m/s^2. Calculate the braking distance.
Final speed v = 0, initial speed u = 20 m/s, a = -4 m/s^2. 0 = 20^2 + 2 x (-4) x s, so 0 = 400 - 8s and s = 400 ÷ 8 = 50 m.
Check with a graph: the speed-time graph is a triangle from 20 m/s to 0 in a time of 20 ÷ 4 = 5 s, and its area is 1/2 x 5 x 20 = 50 m. Getting the same answer two ways is a quick way to catch a sign error. Cambridge students can always use the area method instead of this equation.
Common mistakes that cost marks
- Reading a speed-time graph as if it were a distance-time graph (a horizontal line means constant speed, not stopped).
- Using the area under a distance-time graph. Only the area under a speed-time graph gives distance.
- Calculating the gradient from two points read off a curve, when the question asks for the gradient at one instant. Draw a tangent first.
- Averaging two speeds instead of dividing total distance by total time.
- Leaving out units, or writing m/s for acceleration instead of m/s^2.
- Forgetting to convert minutes to seconds or kilometres to metres before calculating.
- Saying an object at terminal velocity has no forces on it. The forces are balanced, so the resultant force is zero.
Exam technique and mark-scheme language
Calculation questions usually give a mark for the equation or substitution, one for the answer and sometimes one for the unit. Write the equation in symbols, substitute with units, then give the answer to a sensible number of significant figures (usually the same as the data, often 2 or 3). If you make an arithmetic slip, a clear substitution can still earn a method mark.
For "describe the motion" questions, use each section of the graph and give numbers: "from 0 to 4 s the car accelerates uniformly from rest to 12 m/s; from 4 to 10 s it moves at a constant speed of 12 m/s". For terminal velocity, the full chain is: at first weight is greater than air resistance so the object accelerates; as speed increases, air resistance increases; eventually air resistance equals weight, the resultant force is zero and the object falls at constant (terminal) velocity.
How one-to-one lessons help with this topic
Motion is where physics first relies on graph skills from maths, and students who are unsure about gradients struggle quietly. In a lesson a tutor puts past-paper graphs on the shared whiteboard, has the student annotate each section aloud, then sets timed calculations until the equation, substitution and unit routine is automatic. For Edexcel students the tutor adds v^2 = u^2 + 2as problems; for Cambridge Core students the tutor checks tier placement by watching how comfortably they handle Supplement questions.
Self-check: can you do all of these?
- Define speed, velocity and acceleration, with units.
- Explain why velocity is a vector and speed is a scalar.
- Describe the motion shown by each shape on both graph types.
- Find speed from a distance-time gradient and acceleration from a speed-time gradient.
- Find distance from the area under a speed-time graph made of triangles and rectangles.
- Explain terminal velocity in terms of weight and air resistance.
- (Edexcel) Use v^2 = u^2 + 2as to find a stopping distance.
Common questions
Do Cambridge Core students need to calculate acceleration?
In Cambridge 0625, defining acceleration and using a = change in velocity / time is Supplement statement 1.2.9, as is finding acceleration from a gradient. Core students must still recognise accelerating and decelerating motion on graphs and find distance from the area under a speed-time graph.
Is v^2 = u^2 + 2as on the Cambridge syllabus?
No. It is Edexcel 4PH1 statement 1.10. Cambridge students solve the same kinds of problems with the area under a speed-time graph.
What value of g should I use?
The Cambridge syllabus states that the acceleration of free fall near the Earth is approximately 9.8 m/s^2. In any exam, use the value printed in the question if one is given.
How much do LiveTutor physics lessons cost?
$15 a lesson, the same for every subject and level, on a weekly plan of 1 to 5 lessons a week billed monthly. Lessons are 60 minutes, one to one and online, and the first lesson is a free trial.
How do I find the gradient of a curved graph?
Draw a tangent to the curve at the point you need, make a large triangle on the tangent, and divide the change in the vertical value by the change in the horizontal value. Examiners check that the tangent touches the curve at the right point.
Sources
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