AQA GCSE Physics (8463), Higher tier · Forces › Forces and motion › Describing motion along a line
Practise Velocity. 11 exam-style questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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What is meant by the velocity of an object?
Its speed in a given direction.
A parcel on a conveyor belt moves 6.0 m to the right in 4.0 s. Calculate its average velocity.
1.5 m/s to the right m/s
A lift moves upwards at a constant 2.5 m/s. Later, the same lift moves downwards at a constant 2.5 m/s. Compare the lift’s speed in the two cases.
The speed is the same in both cases (2.5 m/s).
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) Which of these is a velocity? Tick (✓) one box.[1]
20 m/s
20 m/s due north
20 m
20 m due north
(b) What is meant by the velocity of an object?[1]
(c) Two trains pass each other on parallel tracks. Each train travels at 30 m/s, but in opposite directions. Explain why the trains have the same speed but different velocities.[2]
Show the answer and mark scheme
(a)Answer: 20 m/s due north
(b)Answer: Its speed in a given direction.
its speed in a given direction
(c)Answer: Speed has no direction so both are 30 m/s; velocity includes direction and the directions are opposite.
speed has no direction, so both trains have a speed of 30 m/s
velocity includes direction, and the trains move in opposite directions
Question 2Medium6 marks
A swimmer swims one length of a 50 m pool in 40 s. The swimmer then turns and swims back to the start in 45 s.
(a) Calculate the average velocity of the swimmer during the first length. Give the direction.[2]
(b) Calculate the average speed of the swimmer for the two lengths.[2]
(c) What is the average velocity of the swimmer for the two lengths? Explain your answer.[2]
Show the answer and mark scheme
(a)Answer: 1.25 m/s away from the start m/s
50 ÷ 40 = 1.25 (m/s)
away from the start / towards the far end of the pool
(b)Answer: 1.2 m/s (1.18 m/s) m/s
100 ÷ 85
1.2 (m/s) / 1.18 (m/s)
(c)Answer: 0 m/s
zero
the swimmer finishes where they started, so the displacement is zero
Question 3Hard7 marks
A ball is rolled up a straight ramp. It slows down, momentarily comes to rest, and then rolls back down the ramp along the same line. The velocity up the ramp is taken as positive.
(a) Describe how the ball’s velocity changes as it travels up the ramp, and give the value of its velocity at the highest point.[2]
(b) Explain why the ball’s velocity is negative while it rolls back down the ramp.[1]
(c) The ball leaves the bottom of the ramp at 3.0 m/s up the ramp. It returns to the bottom 2.4 s later, moving at 3.0 m/s down the ramp. Calculate the size of the change in the ball’s velocity.[2]
(d) Calculate the average acceleration of the ball during the 2.4 s.[2]
Show the answer and mark scheme
(a)Answer: Its velocity decreases up the ramp, reaching zero at the highest point.
its velocity decreases (it decelerates) as it travels up the ramp
at the highest point its velocity is zero
(b)Answer: It is moving in the opposite direction to the direction taken as positive.
velocity is a vector, and the ball is now moving in the opposite direction to the (positive) upward direction
(c)Answer: 6.0 m/s (down the ramp) m/s
change in velocity = (−3.0) − (+3.0) = −6.0 (m/s)
size of the change = 6.0 (m/s) (the change is down the ramp)