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Physics · Forces

Acceleration and velocity–time graphs

Use \(a = \Delta v \div t\), interpret velocity–time graphs, and use \(v^2 - u^2 = 2as\) (and the area under a graph at Higher tier).

  • 6 key terms
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Teacher resources

The teacher copies: slides with the questions built in, the answers, and anything else attached to this lesson for whoever is teaching it.

Student handouts

The same files the students see, to print or hand out.

Warm-up

Answer each one, then check.

  1. 1

    What is velocity?

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    Speed in a given direction

  2. 2

    What is the unit of acceleration?

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    Metres per second squared (m/s²)

  3. 3

    What does deceleration mean?

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    Slowing down

  4. 4

    What is the gradient of a graph?

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    Change in y divided by change in x

  5. 5

    What is the acceleration of free fall near Earth?

    Show answerHide answer

    About 9.8 m/s²

Learning Objectives

  1. 1Recall and apply a = Δv ÷ t.
  2. 2Draw and interpret velocity–time graphs.
  3. 3Use the gradient to find acceleration.
  4. 4Use the area under a velocity–time graph to find distance (Higher tier).
  5. 5Apply \(v^2 - u^2 = 2as\).

ACCELERATION

acceleration \(=\) change in velocity \(\div\) time taken \(a = \dfrac{\Delta v}{t}\)

For uniform acceleration \(v^2 - u^2 = 2as\) (given on the equation sheet). Near Earth's surface a freely falling object accelerates at about 9.8 m/s².

Calculating Acceleration

A car accelerates from rest to 12 m/s in 4.0 s. Calculate its acceleration.

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  1. 1 Change in velocity \(12 - 0 = 12\) m/s
  2. 2 Substitute \(a = 12 \div 4.0\)
  3. 3 Answer \(a = 3.0\) m/s²

Answer3.0 m/s²

Using v² − u² = 2as

A car starts at 10 m/s and accelerates at 2.0 m/s² over a distance of 100 m. Calculate the final speed.

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  1. 1 Write the equation \(v^2 - u^2 = 2as\)
  2. 2 Substitute \(v^2 - 10^2 = 2 \times 2.0 \times 100\)
  3. 3 Rearrange \(v^2 = 100 + 400 = 500\)
  4. 4 Square root \(v = 22.4\) m/s

Answer22 m/s (2 s.f.)

Area Under a Graph (Higher)

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.

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  1. 1 Section 1 (triangle) \(\tfrac{1}{2} \times 5 \times 10 = 25\) m
  2. 2 Section 2 (rectangle) \(10 \times 10 = 100\) m
  3. 3 Section 3 (triangle) \(\tfrac{1}{2} \times 5 \times 10 = 25\) m
  4. 4 Total \(25 + 100 + 25 = 150\) m

Answer150 m

Free Fall

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.

Show the solutionHide the solution
  1. 1 Write the equation \(v^2 = 2as\)
  2. 2 Substitute \(v^2 = 2 \times 9.8 \times 20 = 392\)
  3. 3 Answer \(v = 19.8\) m/s

Answer19.8 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.

Acceleration Practice

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...?

  1. 1Recall a = Δv ÷ t.
  2. 2Calculate acceleration.
  3. 3Use the gradient of a v–t graph.
  4. 4Use the area under a v–t graph.
  5. 5Use \(v^2 - u^2 = 2as\).
  6. 6Estimate everyday accelerations.
  7. 7Give units.
  8. 8Read a graph.

Summary & Exam Focus

  • \(a = \Delta v \div t\).
  • \(v^2 - u^2 = 2as\).
  • v–t gradient = acceleration; area = distance.
  • Free fall: a ≈ 9.8 m/s².

Exam focus

A car accelerates from rest to 12 m/s in 4.0 s. Calculate the acceleration. (2 marks) (2 marks)

Divide the change in velocity by time.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Acceleration
The rate of change of velocity.
Deceleration
Slowing down.
Velocity–time graph
A graph of velocity against time.
Uniform acceleration
Constant acceleration.
Free fall
Falling under gravity alone.
Gradient
How steep a line is.

Questions and answers

10 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 2 marks Easier

A car accelerates from rest to a velocity of 12 m/s in 4.0 s. Calculate the acceleration of the car. Use the equation: acceleration = change in velocity ÷ time taken

Mark scheme — 2 marks available

  • Correct substitution — 1 mark
  • 3.0 m/s² — 1 mark

Model answer

\(12 \div 4.0 = 3.0\) m/s²

2. Exam question Use the graph 4 marks Easier

The graph shows the velocity of a cyclist. (a) Calculate the acceleration in the first 5 s. (b) Describe the motion between 5 s and 15 s.

A velocity-time graph with acceleration to 10 m/s, constant velocity and deceleration to rest.

Mark scheme — 4 marks available

  • Change in velocity 10 m/s — 1 mark
  • 2.0 m/s² — 1 mark
  • Constant velocity — 1 mark
  • 10 m/s — 1 mark

Model answer

(a) \(10 \div 5 = 2.0\) m/s². (b) Constant velocity of 10 m/s (zero acceleration).

3. Exam question Calculate 3 marks Easier

A car starts at 10 m/s and accelerates uniformly at 2.0 m/s² over a distance of 100 m. Calculate its final velocity. Use the equation: (final velocity)² − (initial velocity)² = 2 × acceleration × distance

Mark scheme — 3 marks available

  • Correct substitution — 1 mark
  • v² = 500 — 1 mark
  • 22 m/s — 1 mark

Model answer

\(v^2 = 10^2 + 2 \times 2.0 \times 100 = 500\); \(v = 22\) m/s (22.4)

4. Exam question Calculate 3 marks Easier

Use the velocity–time graph in question 2 to calculate the total distance travelled by the cyclist.

Mark scheme — 3 marks available

  • Areas of the three sections — 1 mark
  • 25 + 100 + 25 — 1 mark
  • 150 m — 1 mark

Model answer

Area = ½ × 5 × 10 + 10 × 10 + ½ × 5 × 10 = 25 + 100 + 25 = 150 m.

5. Exam question Calculate 3 marks Easier

A ball is dropped from rest from a height of 20 m. Assume no air resistance. The acceleration of free fall is 9.8 m/s². Calculate the speed of the ball when it lands.

Mark scheme — 3 marks available

  • Uses v² = 2as — 1 mark
  • 392 — 1 mark
  • 19.8 m/s — 1 mark

Model answer

\(v^2 = 2 \times 9.8 \times 20 = 392\); \(v = 19.8\) m/s

6. Multiple choice 1 mark Easier

The unit of acceleration is...

  1. A m/s² Correct
  2. B m/s
  3. C m
  4. D N

Why: Metres per second squared.

7. Multiple choice 1 mark Core

Acceleration is the gradient of a...

  1. A distance–time graph
  2. B velocity–time graph Correct
  3. C force–time graph
  4. D pie chart

Why: Change in velocity per second.

8. Multiple choice 1 mark Core

From 0 to 20 m/s in 5 s gives...

  1. A 100 m/s²
  2. B 0.25 m/s²
  3. C 4 m/s² Correct
  4. D 25 m/s²

Why: 20 ÷ 5.

9. Multiple choice 1 mark Core

A negative gradient on a v–t graph shows...

  1. A acceleration
  2. B constant velocity
  3. C no motion
  4. D deceleration Correct

Why: The velocity is falling.

10. Multiple choice 1 mark Stretch

Free fall near Earth has acceleration about...

  1. A 9.8 m/s² Correct
  2. B 98 m/s²
  3. C 0.98 m/s²
  4. D 1 m/s²

Why: The acceleration due to gravity.