AQA GCSE PHYSICS · PAPER 2 · FOUNDATION & HIGHER
Acceleration and velocity–time graphs
Forces · Lesson 12 of 20
Warm-up
Answer each one, then check.
1. What is velocity?
Speed in a given direction
2. What is the unit of acceleration?
Metres per second squared (m/s²)
3. What does deceleration mean?
Slowing down
4. What is the gradient of a graph?
Change in y divided by change in x
5. What is the acceleration of free fall near Earth?
About 9.8 m/s²
Learning Objectives
1. Recall and apply a = Δv ÷ t.
2. Draw and interpret velocity–time graphs.
3. Use the gradient to find acceleration.
4. Use the area under a velocity–time graph to find distance (Higher tier).
5. Apply v² − u² = 2as.
Acceleration
acceleration = change in velocity ÷ time taken a = Δv/t
For uniform acceleration v² − u² = 2as (given on the equation sheet). Near Earth's surface a freely falling object accelerates at about 9.8 m/s².
A Velocity–Time Graph
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Gradient = acceleration; area = distance. |
A velocity-time graph showing acceleration, constant velocity and deceleration with the area under it shaded.
Calculating Acceleration
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A car accelerates from rest to 12 m/s in 4.0 s. Calculate its acceleration. |
1. Change in velocity
12 − 0 = 12 m/s
2. Substitute
a = 12 ÷ 4.0
3. Answer
a = 3.0 m/s²
Answer: 3.0 m/s²
Using v² − u² = 2as
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A car starts at 10 m/s and accelerates at 2.0 m/s² over a distance of 100 m. Calculate the final speed. |
1. Write the equation
v² − u² = 2as
2. Substitute
v² − 10² = 2 × 2.0 × 100
3. Rearrange
v² = 100 + 400 = 500
4. Square root
v = 22.4 m/s
Answer: 22 m/s (2 s.f.)
Area Under a Graph (Higher)
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Use the graph (accelerating from 0 to 10 m/s in 5 s, constant for 10 s, then stopping in 5 s) to find the distance travelled. |
1. Section 1 (triangle)
½ × 5 × 10 = 25 m
2. Section 2 (rectangle)
10 × 10 = 100 m
3. Section 3 (triangle)
½ × 5 × 10 = 25 m
4. Total
25 + 100 + 25 = 150 m
Answer: 150 m
Free Fall
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A ball is dropped from rest from a height of 20 m. Ignoring air resistance, calculate its speed on landing. a = 9.8 m/s². |
1. Write the equation
v² = 2as
2. Substitute
v² = 2 × 9.8 × 20 = 392
3. Answer
v = 19.8 m/s
Answer: 19.8 m/s
Reading the Graph
Key features.
▸ Gradient. The acceleration; negative gradient means deceleration.
▸ Horizontal line. Constant velocity (zero acceleration).
▸ Area. Distance travelled (Higher tier); count squares for curves.
▸ Units. m/s for velocity, m/s² for acceleration.
Key Terms
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Acceleration The rate of change of velocity. |
Deceleration Slowing down. |
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Velocity–time graph A graph of velocity against time. |
Uniform acceleration Constant acceleration. |
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Free fall Falling under gravity alone. |
Gradient How steep a line is. |
Your Task: Acceleration Practice
12 minutes
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A train slows from 30 m/s to 10 m/s in 50 s. Calculate its acceleration. What does the sign tell you? 1. Change in velocity. 2. Divide by time. |
A good answer shows: a = (10 − 30) ÷ 50 = −0.40 m/s². The negative value shows deceleration.
Can I...?
☐ Recall a = Δv ÷ t.
☐ Calculate acceleration.
☐ Use the gradient of a v–t graph.
☐ Use the area under a v–t graph.
☐ Use v² − u² = 2as.
☐ Estimate everyday accelerations.
☐ Give units.
☐ Read a graph.
Summary
✓ a = Δv ÷ t.
✓ v² − u² = 2as.
✓ v–t gradient = acceleration; area = distance.
✓ Free fall: a ≈ 9.8 m/s².
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EXAM FOCUS A car accelerates from rest to 12 m/s in 4.0 s. Calculate the acceleration. (2 marks) Divide the change in velocity by time. |