Velocity time graphs Quiz (Grade 8+)

Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Grade 6 Grade 7 Grade 8/9

Topic:

Score: 0 / 10

Velocity time graphs — Notes & Worked Examples Show / Hide

Explanation:

On a velocity-time graph, the gradient represents acceleration (change in velocity ÷ time) and the area under the graph represents distance travelled. For straight-line graphs, split the region into triangles and trapeziums; for curved graphs, draw a tangent at the given point to estimate instantaneous acceleration.

To estimate the distance under a curve using strips of equal width, split the area into a triangle and trapeziums: if the curve is concave down, the chords lie below the curve so your answer is an underestimate; if concave up, it is an overestimate.

Example 1: A car accelerates uniformly from rest to 20 m/s in 8 s, travels at 20 m/s for 15 s, then decelerates to rest in 5 s. Find the deceleration and total distance.

Solution:

Deceleration = 205 = 4 m/s² (gradient = −4 m/s²)

Total time = 8 + 15 + 5 = 28 s (top side = 15 s)

Area of trapezium = 15 + 282 × 20 = 430 m

Answer: 4 m/s² and 430 m

Example 2: Velocity v (m/s) at t = 0, 2, 4, 6 s is 0, 5, 14, 27. Estimate the distance travelled in 6 s using 3 strips of equal width.

Solution:

Strip 1 (0 to 2): 12 × 2 × 5 = 5 m

Strip 2 (2 to 4): 5 + 142 × 2 = 19 m

Strip 3 (4 to 6): 14 + 272 × 2 = 41 m

Answer: 5 + 19 + 41 = 65 m

One Page Guide (PDF) Velocity time graphs Worksheets

Question 1

[2 marks]

The speed-time graph shows the journey of a car.
Calculate the total distance travelled over the 18 seconds.

Worked Solution & Answer

(Distance = speed × time)
Area of trapezium = 12(18 + 8) × 12 = 156 m B1

Answer: 156 m m

Question 2

[2 marks]

The speed-time graph shows the journey of a car.
Calculate the total distance travelled over the 20 seconds.

Worked Solution & Answer

(Distance = speed × time)
Area of trapezium = 12(20 + 10) × 24 = 360 m B1

Answer: 360 m m

Question 3

[3 marks]

The graph shows the speed of a car, in metres per second, during the first 20 seconds of a journey.

(a) Work out an estimate for the distance the car travelled in the first 20 seconds.
Use 4 strips of equal width. [3 marks]

metres

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Strip width = 5 s M1
Speeds at t = 0, 5, 10, 15, 20 are 0, 22, 28, 32, 35 m/s M1
Area = 1/2×5×22 + 1/2×5×(22+28) + 1/2×5×(28+32) + 1/2×5×(32+35) = 55 + 125 + 150 + 167.5 = 497.5 m A1

Answer (a): 497.5 metres

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[3 marks]

Here is a speed-time graph for a car.

(a) Work out an estimate for the distance the car travelled in the first 30 seconds. [2 marks]

(b) Is your answer to part (a) an underestimate or an overestimate of the actual distance the car travelled in the first 30 seconds?
Give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
0.5 × 30 × 9 = 135 m M1 A1

Answer (a): 135 m

(b)

(Average speed = distance ÷ time) C1

Answer (b): Underestimate, because the triangle/trapezium lies below the curve.

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[4 marks]

A car moves from rest.
The graph gives information about the speed, v metres per second, of the car t seconds after it starts to move.

(a) Calculate an estimate of the gradient of the graph at t = 15 [3 marks]

(b) Describe what your answer to part (i) represents. [1 mark]

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Tangent at t = 15 has gradient ≈ 7.5 ÷ 9 = 0.83 M2 A1

Answer (a): 0.83

(b)

(Average speed = distance ÷ time) C1

Answer (b): The acceleration of the car (or rate of change of speed)

Question Paper (PDF) • Mark Scheme (PDF)

Question 6

[4 marks]

Here is a speed-time graph for a train.

(a) Work out an estimate for the distance the train travelled in the first 20 seconds.
Use 4 strips of equal width. [3 marks]

(b) Is your answer to (a) an underestimate or an overestimate of the actual distance the train travelled?
Give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Strip width = 5 s M1
speeds at t = 0, 5, 10, 15, 20 are 0, 2, 5, 10, 18 m/s M1
Area = (0.5 × 5 × 2) + (0.5 × 5 × (2+5)) + (0.5 × 5 × (5+10)) + (0.5 × 5 × (10+18)) = 5 + 17.5 + 37.5 + 70 = 130 m A1

Answer (a): 130 m

(b)

(Average speed = distance ÷ time)
Overestimate since chords lie above the curve C1

Answer (b): Overestimate, because the tops of the trapeziums lie above the curve (the curve is convex)

Question Paper (PDF) • Mark Scheme (PDF)

Question 7

[4 marks]

Here is a speed-time graph showing the speed, in metres per second, of an object t seconds after it started to move from rest.

(a) Using 3 trapeziums of equal width, work out an estimate for the area under the graph between t = 1 and t = 4 [3 marks]

(b) What does this area represent? [1 mark]

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Ordinates at t = 1, 2, 3, 4 are 4, 6, 7.2, 7.8. Area = ½(1)(4 + 6) + ½(1)(6 + 7.2) + ½(1)(7.2 + 7.8) = 5 + 6.6 + 7.5 = 19.1 M2 A1

Answer (a): 19.1

(b)

(Average speed = distance ÷ time)
Distance travelled in metres B1

Answer (b): Distance travelled (between t = 1 and t = 4 seconds)

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[4 marks]

Here is a speed-time graph.

(a) Work out an estimate of the gradient of the graph at t = 2 [3 marks]

(b) What does the area under the graph represent? [1 mark]

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Draw tangent at t = 2: change in speed ÷ change in time ≈ 2.7 ÷ 3 = 0.9 M2 A1

Answer (a): 0.9

(b)

(Average speed = distance ÷ time)
Area under a speed-time graph represents distance travelled C1

Answer (b): Distance travelled

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 AQA Nov 2023 2H Q23

[5 marks]

Here is the velocity-time graph of a cyclist for 40 seconds.

(a) By dividing the area under the graph into four sections of equal widths,
estimate the distance travelled during the 40 seconds. [3 marks]

Answer m

(b) Work out the average acceleration of the cyclist during the 40 seconds.
State the units of your answer. [2 marks]

Answer

Worked Solution & Answer

(a)

Velocities at t = 0, 10, 20, 30, 40 are 0, 3.2, 6.0, 7.4, 6.0 m/s M1
Areas of 4 strips of width 10:
Strip 1: 1/2 × 10 × 3.2 = 16
Strip 2: 1/2 × (3.2 + 6.0) × 10 = 46 M1
Strip 3: 1/2 × (6.0 + 7.4) × 10 = 67
Strip 4: 1/2 × (7.4 + 6.0) × 10 = 67
Total distance = 16 + 46 + 67 + 67 = 196 m A1

Answer (a): Answer 196 m

(b)

Average acceleration = (6.0 – 0) ÷ 40 = 0.15 m/s² M1 A1

Answer (b): Answer 0.15 m/s²

Question Paper (PDF) • Mark Scheme (PDF)

Question 10

[5 marks]

Here is a speed-time graph for a train journey between two stations.
The journey took 100 seconds.

(a) Calculate the time taken by the train to travel half the distance between the two stations.
You must show all your working. [4 marks]

seconds

(b) Compare the acceleration of the train during the first part of its journey with the acceleration of the train during the last part of its journey. [1 mark]

Worked Solution & Answer

(a)

(Time = distance ÷ speed)
Total distance = ½(30×12) + (50×12) + ½(20×12) = 180 + 600 + 120 = 900 m M2
Half distance = 450 m M1
Time = 30 + (450 – 180) ÷ 12 = 30 + 22.5 = 52.5 s A1

Answer (a): 52.5 seconds

(b)

(Average speed = distance ÷ time) C1

Answer (b): The acceleration during the first part is positive (0.4 m/s²), whereas during the last part it is negative / deceleration (–0.6 m/s², or steeper magnitude).

Question Paper (PDF) • Mark Scheme (PDF)

Score: 0 / 10 (0 / 36 marks)

Velocity time graphs Worksheets Next: Circles and tangents →

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum