Step-by-step worked examples and graded practice questions on velocity-time graphs — reading acceleration from the gradient, finding distance from the area under the graph, and multi-stage journeys.
📚 Foundation & Higher✅ 15 Practice Questions🔍 Full Worked Examples⚠️ Common Mistakes
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A velocity-time graph shows velocity plotted against time. Unlike a distance-time graph, it carries two separate pieces of information:
The gradient of the line represents acceleration
The area under the graph represents distance travelled
A horizontal line means constant velocity (zero acceleration). A line sloping upward means the object is speeding up (accelerating); a line sloping downward means it is slowing down (decelerating).
A typical journey: accelerating, then constant velocity, then decelerating to a stop.
Gradient = acceleration
Acceleration is calculated in exactly the same way as gradient: acceleration = change in velocity ÷ change in time, measured in m/s². A negative gradient means the object is decelerating (slowing down).
Worked Example 1 — Acceleration from a straight line
A car's velocity-time graph is a straight line from (0, 0) to (5, 20), where x is in seconds and y is in m/s. Find the car's acceleration.
1
Acceleration = gradient = change in velocity ÷ change in time
2
Acceleration = (20 − 0) / (5 − 0) = 4 m/s²
Answer4 m/s²
Area under the graph = distance
The area between the line and the time-axis gives the total distance travelled. Split the shape into triangles, rectangles or trapeziums depending on the shape of the graph, and use the standard area formulas.
Worked Example 2 — Distance from the area under the graph
Using the graph from Worked Example 1, find the distance travelled during the 5 seconds of acceleration.
1
The area under the line is a triangle, with base 5 (seconds) and height 20 (m/s)
2
Area = ½ × base × height = ½ × 5 × 20 = 50 m
Answer50 m
The shaded triangle is the area under the graph — the distance travelled during acceleration (50 m).
Multi-stage velocity-time graphs
Many exam questions combine acceleration, constant velocity and deceleration in a single graph — this shape is a trapezium. Find the total distance by adding the areas of each individual section (usually a triangle, a rectangle, and another triangle).
Worked Example 3 — A three-stage journey
A car accelerates from rest to 12 m/s in 4 seconds, travels at a constant 12 m/s for 6 seconds, then decelerates to rest in 3 seconds. Find: (a) the acceleration during the first stage, (b) the deceleration during the final stage, and (c) the total distance travelled.
1
(a) Acceleration = 12 / 4 = 3 m/s²
2
(b) Deceleration: gradient = (0 − 12) / 3 = −4, so the deceleration is 4 m/s²
3
(c) Stage 1 (triangle): ½ × 4 × 12 = 24 m. Stage 2 (rectangle): 12 × 6 = 72 m. Stage 3 (triangle): ½ × 3 × 12 = 18 m
4
Total distance = 24 + 72 + 18 = 114 m
Answer(a) 3 m/s² (b) 4 m/s² (c) 114 m
The trapezium shape: accelerate (0–4s), constant velocity (4–10s), decelerate (10–13s). Shaded area = total distance (114 m).
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1A cyclist's velocity-time graph is a straight line from (0, 0) to (4, 8), where x is in seconds and y is in m/s. Find the acceleration.Foundation
Q2A car travels at a constant velocity of 10 m/s for 6 seconds. Find the distance travelled.Foundation
Q3On a velocity-time graph, what does a horizontal line represent?Foundation
Q4A runner accelerates from 0 to 6 m/s in 3 seconds. Find the acceleration.Foundation
Q5A car decelerates from 20 m/s to 0 m/s in 5 seconds. Find the deceleration.Foundation
Higher (Grade 5–7)
Q6A cyclist's velocity-time graph is a straight line from (0, 3) to (5, 13). Find the acceleration.Higher
Q7Find the distance travelled by the cyclist in Q6 (the area under the graph is a trapezium).Higher
Q8A car travels at a constant 18 m/s for 8 seconds, then decelerates to rest in 3 seconds. Find the total distance travelled.Higher
Q9Using Q8, find the deceleration during the final stage.Higher
Q10A velocity-time graph shows acceleration from 0 to 10 m/s in 2 seconds, constant velocity for 4 seconds, then deceleration to 0 in 2 seconds. Find the total distance travelled.Higher
Higher — Hard (Grade 8–9)
Question 11 refers to the graph below.
Graph for Question 11 — the shaded area is the distance travelled.
Q11Using the graph above, find (a) the acceleration and (b) the distance travelled.Grade 8–9
Q12A car accelerates uniformly from rest to V m/s in 8 seconds, covering 96 m. Find V.Grade 8–9
Q13A train decelerates uniformly from 40 m/s to 10 m/s over 6 seconds. Find (a) the deceleration and (b) the distance travelled during deceleration.Grade 8–9
Q14A cyclist accelerates from rest to 8 m/s in 4 seconds, maintains this speed for 10 seconds, then decelerates to rest, covering a total distance of 140 m. Find the time taken to decelerate.Grade 8–9
Q15A velocity-time graph is a straight line from (0, u) to (10, v) with constant acceleration 3 m/s², and the area under the graph between t = 0 and t = 10 is 250 m. Find u and v.Grade 8–9
Q15u = 10 m/s, v = 40 m/s(v = u+30 from a=3 over 10s; ½(u+v)×10=250 → u+v=50 → u=10, v=40)
Common mistakes
These are the errors Alamin sees most frequently with velocity-time graphs at GCSE. Recognising them now will save you marks in the exam.
Common Mistake 1
Confusing velocity-time graphs with distance-time graphs
On a velocity-time graph, the gradient is acceleration, not speed. Students who treat it like a distance-time graph read the wrong quantity entirely.
Common Mistake 2
Forgetting that area gives distance
A common error is trying to read distance travelled directly off the velocity axis. Distance is always the area under the graph, not a y-value.
Common Mistake 3
Using the wrong area formula for the shape
Check whether each section is a triangle (½ × base × height), a rectangle (base × height), or a trapezium (½ × (a+b) × height) before calculating — using the wrong formula gives a wrong area every time.
Common Mistake 4
Giving a negative deceleration as the final answer
A negative gradient means the object is decelerating. State the deceleration as a positive number (e.g. "deceleration of 4 m/s²"), rather than leaving the answer as −4 m/s².
Common Mistake 5
Adding areas incorrectly in multi-stage graphs
Calculate the area of each section separately, then add them together at the end. Trying to use one combined formula for the whole trapezium shape often leads to mistakes — break it down stage by stage.
Common Mistake 6
Mixing up acceleration units
Acceleration is measured in m/s² (metres per second, per second) — not m/s. Always double-check the units requested match what the graph's axes are actually measuring.
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