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Displacement–time and velocity–time graphs for motion in a straight line: what the gradient and the area mean and how to use them. Expect to sketch a velocity–time graph from a description, then use its gradients and areas to find an unknown time, speed or distance.
Key facts
Displacement–time graph: gradient = velocity
Velocity–time graph: gradient = acceleration
Velocity–time graph: area = displacement
Distance = total area, with every part counted as positive
Trapezium area \(= \dfrac{1}{2}(a + b)h\)
Lines crossing on a velocity–time graph: equal velocities, not the same position
Notes
Displacement–time graphs
diagram
The gradient is the velocity. A steeper line means a greater speed.
A horizontal line means the object is at rest. A straight sloping line means constant velocity. A curve means the velocity is changing.
A negative gradient means the object is moving in the negative direction, for example back towards the start.
Where the graph meets the time axis, the object is back at its starting point (displacement zero).
Displacement–time: the gradient is the velocity
Velocity–time graphs
diagram
The gradient is the acceleration; a negative gradient is a deceleration.
The area between the graph and the time axis is the displacement.
A horizontal line means constant velocity; a straight sloping line means constant acceleration.
Below the time axis the velocity is negative: the object moves in the negative direction and that area counts as negative displacement.
Distance travelled = total area, counting every part as positive. Displacement = area above the axis − area below it.
Velocity–time: gradient = acceleration, area = displacement (below the axis counts as negative)
Areas you need: triangle \(\dfrac{1}{2}bh\), rectangle \(bh\), trapezium \(\dfrac{1}{2}(a + b)h\)
Solving problems with graphs
Sketch the velocity–time graph first, even if the question does not ask for it. Label speeds and times, using letters such as \(T\) or \(V\) for unknowns.
Form equations from gradients (accelerations) and areas (distances), then solve.
The time to reach speed \(V\) from rest with acceleration \(a\) is \(\dfrac{V}{a}\)
For two objects, draw both graphs on the same axes. They are level when their displacements from the same starting point are equal. Where two velocity–time lines cross, the objects have equal velocities, not equal positions.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Sketch a velocity–time graph and find the distance
3 to 5 marksGrade E
Split the journey into stages; each constant-acceleration stage is a straight line.
Label the axes (\(v\) in m s−1, \(t\) in s) and mark every key speed and time.
Find the distance as the area, split into triangles and rectangles or treated as a trapezium.
Example. A car starts from rest and accelerates uniformly to 12 m s−1 in 8 s. It travels at 12 m s−1 for 30 s and then decelerates uniformly, coming to rest after a further 6 s. (a) Sketch a velocity–time graph for the motion. (b) Find the total distance travelled by the car.
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(a) As in the diagram: a trapezium shape starting and finishing on the time axis (B1), with 12, 8, 38 and 44 marked
(B1)
(b) Area \(= \dfrac{1}{2}(30 + 44) \times 12\) (M1) = 444 m
(A1)
Interpret a displacement–time graph
3 to 6 marksGrade D
Find the gradient of each straight section: that is the velocity.
A horizontal section means at rest; a negative gradient means moving back.
Average speed = total distance ÷ total time; average velocity = final displacement ÷ total time.
Example. The displacement–time graph for a runner is made of three straight line segments joining the points (0, 0), (20, 100), (50, 100) and (90, 0), where \(t\) is the time in seconds and \(s\) is the displacement in metres from the start.
(a) Find the velocity of the runner in each of the three stages. (b) Find the average speed of the runner over the 90 seconds.
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(a) \(\dfrac{100}{20}\) = 5 m s−1 (B1); 0 m s−1, at rest (B1); \(\dfrac{0 - 100}{90 - 50} = -2.5\) m s−1, i.e. 2.5 m s−1 back towards the start
Sketch the graph and find the time for each accelerating or decelerating stage, using gradient = acceleration.
Write the total area in terms of the unknown.
Set the area equal to the total distance, in metres, and solve.
Example. A train travels between two stations that are 3 km apart. It starts from rest and accelerates uniformly at 0.5 m s−2 until its speed is 20 m s−1. It travels at 20 m s−1 for \(T\) seconds and then decelerates uniformly at 0.8 m s−2, coming to rest at the second station. Find the value of \(T\)
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Time accelerating \(= \dfrac{20}{0.5} = 40\) s (B1); time decelerating \(= \dfrac{20}{0.8} = 25\) s
Sketch both velocity–time graphs on the same axes.
Write each vehicle's displacement (area) at time \(t\), checking which stage each vehicle is in.
Set the displacements equal and solve; check that the time is in the stage you assumed.
Example. A car is waiting at traffic lights. When the lights turn green, the car accelerates uniformly at 2 m s−2 until it reaches a speed of 16 m s−1, and then continues at that speed. At the moment the car starts, a lorry passes it, travelling at a constant 12 m s−1 in the same direction. Find the time taken for the car to catch up with the lorry.
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The car reaches 16 m s−1 after \(\dfrac{16}{2} = 8\) s, having travelled \(\dfrac{1}{2} \times 8 \times 16 = 64\) m; the lorry has then travelled 96 m, so the car is still behind
\(64 + 16(t - 8) = 12t\) (M1), so \(4t = 64\) and \(t = 16\) s
(A1)
Don’t lose marks
Thinking that where two velocity–time lines cross, one object overtakes the other.
Using the area under a displacement–time graph: it has no meaning here.
Subtracting the area below the axis when the question asks for the distance travelled.
Sketches without labels or key values: the B marks are lost.
Assuming a vehicle is still accelerating at the time found, without checking which stage it is in.
More tips
Memory tricks
Rest-to-rest trapezium graph: distance \(= \dfrac{1}{2}\) × (time at top speed + total time) × top speed.
Speed \(V\) reached from rest with acceleration \(a\): time \(\dfrac{V}{a}\), distance \(\dfrac{V^2}{2a}\)
Check your answer: the total area of your sketch should equal the distance given in the question.
Convert km to m before equating an area (in m) to a distance.
Exam technique
A sketch needs the correct shape, labelled axes and the key values marked; it does not need to be to scale.
Use the letters from the question (\(T\),\(V\)) on your sketch and in your equations.
If the question says 'use your graph', form equations from gradients and areas rather than using suvat.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Find velocity from a displacement–time graphThe gradient of a displacement–time graph is the velocity.
Grade E
Sketch a velocity–time graph from a descriptionUse straight lines for constant acceleration and label the key speeds and times.
Grade D
Find acceleration and distance from a velocity–time graphThe gradient gives the acceleration and the area under the graph gives the displacement.
Grade C
Use area to find an unknown timeWrite the total area in terms of the unknown and set it equal to the given distance.
Grade B
Compare two objects on one graphEqual areas from the same start mean the objects are level; lines crossing only means equal velocities.
Grade A
Handle graphs with negative velocityArea below the time axis is negative displacement; for the distance, add the sizes of all the areas.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A cyclist starts from rest and accelerates uniformly to a speed of 12 m s−1 in 4 s. The cyclist then travels at 12 m s−1 for 10 s, and then decelerates uniformly to rest in 6 s. The velocity–time graph shows the motion.
[object Object]
(a) A student says that the distance travelled in the first 4 seconds is \(12 \div 4 = 3\) m. Explain the student’s error and find the correct distance.[2]
(b) Find the total distance travelled by the cyclist.[2]
(c) Find the deceleration of the cyclist in the final 6 seconds.[1]
Show the answer and mark scheme
(a)Answer: 24 m
B1 for \(12 \div 4\) is the gradient, which gives the acceleration (3 m s−2), not the distance; the distance is the area under the graph
B1 for 24 (m)
Worked solution: \(12 \div 4 = 3\) is the gradient of the graph, which is the acceleration (3 m s−2). Distance is the area under the graph: \(\frac{1}{2} \times 4 \times 12 = 24\) m.
(b)Answer: 180 m
M1 for the area of the trapezium, e.g. \(\frac{1}{2}(10 + 20) \times 12\)
A1 for 180 (m)
Worked solution: Area \(= \frac{1}{2}(10 + 20) \times 12 = 180\) m (or \(24 + 120 + 36\)).
(c)Answer: 2 m s−2
B1 for 2 (m s−2)
Worked solution: Deceleration \(= \frac{12}{6} = 2\) m s−2.
Question 2Medium8 marks
A train moves along a straight horizontal track. Starting from rest at the point \(A\), the train accelerates uniformly for \(2T\) seconds, reaching a speed of 20 m s−1. The train then travels at this speed for \(5T\) seconds, and then decelerates uniformly for \(T\) seconds, coming to rest at the point \(B\). The distance \(AB\) is 5200 m.
(a) Sketch a speed–time graph for the motion of the train from \(A\) to \(B\).[2]
(b) Find the value of \(T\).[4]
(c) Find the acceleration of the train during the first stage of the motion.[2]
Show the answer and mark scheme
(a)Answer: A trapezium: a straight line from \((0, 0)\) up to \((2T, 20)\), horizontal to \((7T, 20)\), then a straight line down to \((8T, 0)\).
B1 for a trapezium shape starting at the origin and finishing on the \(t\)-axis
B1 for 20, \(2T\), \(7T\) and \(8T\) marked correctly
Worked solution: The graph rises in a straight line from the origin to speed 20 at \(t = 2T\), is horizontal until \(t = 7T\), and then falls in a straight line to zero at \(t = 8T\).
(b)Answer: \(T = 40\)
M1 for using the area under their graph to form an equation in \(T\)
A1 for a correct equation, e.g. \(\tfrac{1}{2}(5T + 8T) \times 20 = 5200\)
dM1 for solving their linear equation
A1 for \(T = 40\)
Worked solution: Area of the trapezium = distance: \(\tfrac{1}{2}(5T + 8T) \times 20 = 5200\) \(130T = 5200 \Rightarrow T = 40\)
(c)Answer: \(0.25\) m s−2
M1 for \(\frac{20}{2 \times \text{their }T}\)
A1ft for 0.25 (m s−2)
Worked solution: \(a = \frac{20}{2 \times 40} = 0.25\text{ m s}^{-2}\)
Question 3Hard12 marks
Two cars, \(A\) and \(B\), move along a straight horizontal road in the same direction. At time \(t = 0\), car \(A\) starts from rest at the point \(O\) and accelerates uniformly at 2 m s−2 until it reaches a speed of 24 m s−1. It then moves at this constant speed. Also at \(t = 0\), car \(B\) passes \(O\) moving at a constant speed of 18 m s−1.
(a) Sketch, on the same axes, the velocity–time graphs for the two cars, for the interval from \(t = 0\) until car \(A\) overtakes car \(B\).[3]
(b) Find the time at which car \(A\) overtakes car \(B\).[5]
(c) Find the distance from \(O\) at which car \(A\) overtakes car \(B\).[1]
(d) Find the greatest distance between the two cars before car \(A\) overtakes car \(B\).[3]
Show the answer and mark scheme
(a)Answer: \(A\): a straight line from the origin to \((12, 24)\), then horizontal at 24. \(B\): a horizontal line at 18, starting on the velocity axis. The lines cross at \(t = 9\).
B1 for the graph of \(A\): a line through the origin then horizontal
B1 for the graph of \(B\): a horizontal line starting on the velocity axis, below the horizontal part of \(A\)’s graph
B1 for 24 and 18 marked on the velocity axis and 12 on the time axis
Worked solution: Car \(A\) reaches 24 m s−1 at \(t = \frac{24}{2} = 12\). Its graph is a straight line from the origin to \((12, 24)\) and then horizontal. Car \(B\)’s graph is a horizontal line at 18.
(b)Answer: \(t = 24\) s
M1 for the distance travelled by \(B\) as \(18t\)
M1 for the distance travelled by \(A\) as the area under its graph, e.g. \(\tfrac{1}{2} \times 12 \times 24 + 24(t - 12)\)
M1 for equating the two distances
A1 for a correct equation, e.g. \(24t - 144 = 18t\)
A1 for \(t = 24\)
Worked solution: \(A\) reaches its top speed at \(t = 12\), having travelled \(\tfrac{1}{2} \times 12 \times 24 = 144\) m, while \(B\) has travelled \(216\) m; so \(A\) has not yet caught up. For \(t \gt 12\): \(144 + 24(t - 12) = 18t \Rightarrow 6t = 144 \Rightarrow t = 24\)
(c)Answer: 432 m
B1ft for 432 (m)
Worked solution: \(18 \times 24 = 432\) m
(d)Answer: \(81\) m
M1 for recognising that the gap is greatest when the two speeds are equal, at \(t = 9\)
M1 for the difference in the areas under the two graphs up to this time
A1 for 81 (m)
Worked solution: The gap grows while \(B\) is faster and shrinks once \(A\) is faster, so it is greatest when both speeds equal 18 m s−1, at \(t = \frac{18}{2} = 9\). Gap = \(18 \times 9 - \tfrac{1}{2} \times 9 \times 18 = 81\) m (the area of the triangle between the two graphs).