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Speed Quiz

Grade 5 • 5E • 10 Questions (35 Marks)

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Speed — Notes & Worked Examples Show / Hide

Explanation:

Speed, distance and time are connected by Speed = DistanceTime, Distance = Speed × Time, and Time = DistanceSpeed. Always check that your units match: to convert minutes into hours, divide by 60 (for example, 45 minutes = 4560 = 0.75 hours, never 0.45).

For a journey in two or more stages, never average the speeds—always use Average Speed = Total DistanceTotal Time.

Example 1: A train travels 198 km in 2 hours 15 minutes. Work out its average speed in km/h.

Solution:

Convert time to hours: 15 ÷ 60 = 0.25 hours
Total time = 2.25 hours
Speed = Distance ÷ Time = 198 ÷ 2.25
198 ÷ 2.25 = 88
Answer: 88 km/h

Example 2: Ben drives 60 miles at 40 mph, then 90 miles at 60 mph. Find his average speed for the whole journey.

Solution:

Time 1 = 60 ÷ 40 = 1.5 h; Time 2 = 90 ÷ 60 = 1.5 h
Total distance = 60 + 90 = 150 miles
Total time = 1.5 + 1.5 = 3 hours
Average speed = 150 ÷ 3 = 50 mph
Answer: 50 mph
Question 1
[1 mark]
Work out
A car travels 100 miles in 2 hours and 30 minutes
Calculate the average speed of the car in mph
mph
Worked Solution & Answer
(Average speed = distance ÷ time) B1
Answer: 40 mph mph
Question 2
[1 mark]
Work out
A train travels at an average speed of 50 km/h
It travels a distance of 112.5 km
Work out how long the journey takes in hours and minutes
Worked Solution & Answer
(Time = distance ÷ speed) B1
Answer: 2h 15m
Question 3 OCR Jun 2017 2F Q15 / 5H Q2
[3 marks]
The graph shows Mia’s journey from her home to university.
020406080100120140160090010001100TimeuniversityhomeDistancefrom home(kilometres)
Calculate Mia’s average speed for the whole journey.
km/h
Worked Solution & Answer
(Average speed = distance ÷ time)
Total distance = 160 km M1
Total time from 0900 to 1130 is 2.5 hours. M1
Average speed = 1602.5 = 64 km/h. A1
Answer: 64 km/h km/h
Question Paper (PDF) • Mark Scheme (PDF)
Question 4 AQA Nov 2021 1F Q28 / 1H Q11
[4 marks]
A, B, C and D are junctions on a motorway.
Not drawnaccuratelyABCD
distance CD = 3 × distance AB
distance BC = 25 miles
Salma drives from A to C.
She drives for 30 minutes at an average speed of 62 miles per hour.
Work out the distance AD.
Answer miles
Worked Solution & Answer
(Distance = speed × time)
Time from A to C = 0.5 hours M1
Distance AC = 62 × 0.5 = 31 miles M1
Distance AB = AC - BC = 31 - 25 = 6 miles M1
Distance CD = 3 × AB = 3 × 6 = 18 miles
Distance AD = AC + CD = 31 + 18 = 49 miles. A1
Answer: Answer 49 miles
Question Paper (PDF) • Mark Scheme (PDF)
Question 5
[4 marks]
Nimer was driving to a hotel.
He looked at his Sat Nav at 13 30
Time13 30Distance to destination65 miles
Nimer arrived at the hotel at 14 48
Work out the average speed of the car from 13 30 to 14 48
You must show all your working.
mph
Worked Solution & Answer
(Average speed = distance ÷ time)
Time from 13 30 to 14 48 = 1 hour 18 minutes = 78 minutes = 1.3 hours M3
Average speed = 65 ÷ 1.3 = 50 mph A1
Answer: 50 mph
Question Paper (PDF) • Mark Scheme (PDF)
Question 6 Eduqas Jun 2024 2F Q9
[4 marks]
The table below shows distances between some places in New Zealand.
All distances are in kilometres (km).
New Plymouth357797Rotorua234670290Thames114550414349Wellington6581094718323380Whakatane298737205308165AucklandCape ReingaHicks BayNapierTaupo
(a) Complete the sentence. [1 mark]
(b) Sharon drove from Wellington to Taupo.
The journey took 4 hours and 30 minutes.
Calculate the average speed of Sharon’s journey in kilometres per hour. [3 marks]
km/h
Worked Solution & Answer
(a)
(Average speed = distance ÷ time)
Look up 414 in the table: row Thames and column Hicks Bay. B1
Answer (a): Thames and Hicks Bay
(b)
(Average speed = distance ÷ time)
Distance from Wellington to Taupo = 380 km M1
Time = 4 hours 30 minutes = 4· 5 hours. M1
Average speed = 3804· 5 = 84· 444... km/h ≈ 84· 4 km/h. A1
Answer (b): 84.4 km/h
Question Paper (PDF) • Mark Scheme (PDF)
Question 7 AQA Jun 2017 2F Q23 / 2H Q8
[4 marks]
Lily goes on a car journey.
For the first 30 minutes her average speed is 40 miles per hour.
She then stops for 15 minutes.
She then completes the journey at an average speed of 60 miles per hour.
The total journey time is 1 hour.
051015202530354045505560510152025303540Time taken (minutes)Distance travelled (miles)
(a) Draw a distance-time graph for her journey. [3 marks]
(b) Write down the average speed for the total journey. [1 mark]
Answer mph
Worked Solution & Answer
(a)
(Average speed = distance ÷ time)
First leg: 30 minutes = 0.5 h M1
Distance = 40 × 0.5 = 20 miles
Stop: from 30 to 45 min, distance remains 20 miles M1
Final leg: time remaining = 60 - 45 = 15 min = 0.25 h
Distance = 60 × 0.25 = 15 miles
Total distance = 20 + 15 = 35 miles at t = 60 min. A1
Answer (a): Graph drawn: (0, 0) to (30, 20), horizontal to (45, 20), then straight to (60, 35)
(b)
(Average speed = distance ÷ time)
Average speed = total distancetotal time = 35 miles1 hour = 35 mph B1
Answer (b): Answer 35 mph
Question Paper (PDF) • Mark Scheme (PDF)
Question 8 OCR Nov 2022 5H Q16
[4 marks]
The graph shows the distance travelled by a particle over the first 20 seconds of its motion.
05101520510152025303540Time (seconds)Distance(metres)
(a) Show that the average speed of the particle over the first 20 seconds of its motion is 1.8 m/s. [1 mark]
(b) Estimate the speed of the particle at 10 seconds.
You must show working to support your estimate. [3 marks]
m/s
Worked Solution & Answer
(a)
(Average speed = distance ÷ time)
Total distance at t = 20 s is 36 m
Average speed = 36 ÷ 20 = 1.8 m/s. C1
Answer (a): 36 ÷ 20 = 1.8 m/s.
(b)
(Average speed = distance ÷ time)
Draw a tangent to the curve at t = 10 s (passing through (0, 17.5) and (18, 40)) M2
Gradient = (40 − 17.5) ÷ (18 − 0) = 22.5 ÷ 18 = 1.25 m/s (or 1.0 to 1.5). A1
Answer (b): 1.25 m/s
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 Eduqas Nov 2020 2F Q15
[5 marks]
Nadia is going to drive from her home to a meeting in Newcastle.
Her route can be approximated using a straight line.
She plans to leave home at 6 a.m.
She wants to arrive in Newcastle at 11:45 a.m.
Scale: 1 cm represents 50 milesNewcastleHome
What must Nadia’s average speed be for her to get to Newcastle on time?
Give your answer in miles per hour.
mph
Worked Solution & Answer
(Average speed = distance ÷ time)
Time taken = 6:00 to 11:45 = 5.75 hours M2
Map distance = 5.75 cm (represented as 5.75 × 50 = 287.5 miles) M2
Average speed = 287.5 / 5.75 = 50 mph (or 46 to 52 mph depending on map measurement). A1
Answer: 50 mph mph
Question Paper (PDF) • Mark Scheme (PDF)
Question 10 Eduqas Nov 2023 2F Q15
[5 marks]
Rhiannon cycled from her home to a park.
She cycled 15 miles along a straight road.
The graph shows her journey.
Distance from Rhiannon's home (miles)051015202509:0009:3010:0010:3011:0011:3012:0012:3013:00Time
(a) On her journey to the park, Rhiannon had a puncture.
How long did Rhiannon stop to repair the puncture? [1 mark]
(b) Rhiannon was, on average, travelling faster before she had a puncture than after she had a puncture.
Without calculating any speeds, explain how you can tell this from the graph. [1 mark]
(c) Rhiannon left the park at 11:30 and cycled back along the same road to a friend’s house that is 5 miles from her home.
This journey took 45 minutes.
Draw this journey on the graph. [1 mark]
(d) Calculate the average speed of Rhiannon’s journey from the park to her friend’s house.
Give your answer in miles per hour. [2 marks]
Worked Solution & Answer
(a)
(Time = distance ÷ speed)
Flat section from 09:36 to 10:00 is 24 minutes. C1
Answer (a): 24 minutes (allow 20 to 25 minutes)
(b)
(Average speed = distance ÷ time)
Greater gradient on a distance-time graph represents higher speed. C1
Answer (b): The line is steeper before the puncture than after.
(c)
(Average speed = distance ÷ time)
11:30 + 45 mins = 12:15
Connect (11:30, 15) to (12:15, 5). B1
Answer (c): Straight line drawn from (11:30, 15) to (12:15, 5)
(d)
(Average speed = distance ÷ time)
Distance = 15 - 5 = 10 miles M1
Time = 45 mins = 0.75 hours
Speed = 10 / 0.75 = 13⅓ mph. A1
Answer (d): 13⅓ mph (or 13.33 mph)
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 35 marks)
Speed Worksheets Next: Changing the subject →

Grade 5 Quizzes

25 Topics
5A - Combining ratios5B - Direct and inverse proportion
5C - Reverse percentages5D - Standard form
5E - Speed5F - Changing the subject
5G - Quadratic expressions5H - Solving quadratic equations
5J - Drawing quadratic graphs5K - Cubic/reciprocal graphs
5L - Simultaneous equations5M - Sector areas and arc lengths
5N - Sim. equ. graphically5P - Midpoint of a line segment
5Q - Gradient of a line5R - Equation of a line
5S - Spheres and cones5T - Similar shapes (lengths)
5U - Trigonometry (SOHCAHTOA)5V - Exact trig. values
5W - Probability trees5X - Vectors
5Y - Venn diagrams5Z - Density
5AA - Pressure