Motion
Cambridge IGCSE Physics (0625) · Unit 1: Motion, forces and energy · 26 flashcards
Motion is topic 1.2 in the Cambridge IGCSE Physics (0625) syllabus , positioned in Unit 1 — Motion, forces and energy , alongside Physical quantities and measurement techniques, Mass and weight and Density. In one line: Velocity is speed in a given direction. It is a vector quantity, meaning it has both magnitude (speed) and direction. (2 marks).
This topic is examined in Paper 1 (multiple-choice) and Papers 3/4 (theory), plus Paper 5 or Paper 6 (practical / alternative to practical). Past papers from 2022 to 2025 show this topic across undefined questions worth 362 marks (around 5.7% of all Physics marks in those years).
The deck below contains 26 flashcards — 2 definitions — covering the precise wording mark schemes reward. Use the 2 definition cards to lock down command-word answers (define, state), then move on to the concept and application cards to handle explain, describe and compare questions.
Velocity
Velocity is speed in a given direction. It is a vector quantity, meaning it has both magnitude (speed) and direction. (2 marks)
What the Cambridge 0625 syllabus says
Official 2026-2028 specThese are the exact learning objectives Cambridge sets for this topic. Match the command word (Describe, Explain, State, etc.) in your answer to score full marks.
- Define Define speed as distance travelled per unit time; recall and use the equation
- Define Define velocity as speed in a given direction
- Recall Recall and use the equation average speed =
- Sketch Sketch, plot and interpret distance-time and speed-time graphs
- Determine Determine, qualitatively, from given data or the shape of a distance-time graph or speed-time graph when an object is: (a) at rest (b) moving with constant speed (c) accelerating (d) decelerating
- Calculate Calculate speed from the gradient of a straight-line section of a distance-time graph
- Calculate Calculate the area under a speed-time graph to determine the distance travelled for motion with constant speed or constant acceleration
- State State that the acceleration of free fall g for an object near to the surface of the Earth is approximately constant and is approximately 9.8m/s²
- Define Define acceleration as change in velocity per unit time; recall and use the equation Supplement
- Determine Determine from given data or the shape of a speed-time graph when an object is moving with: (a) constant acceleration (b) changing acceleration Supplement
- Calculate Calculate acceleration from the gradient of a speed-time graph Supplement
- Know Know that a deceleration is a negative acceleration and use this in calculations Supplement
- Describe Describe the motion of objects falling in a uniform gravitational field with and without air/liquid resistance, including reference to terminal velocity Supplement
Define speed. A cyclist travels 50 meters in 5 seconds. Calculate their average speed.
Definition: Speed is the distance travelled per unit time.
Calculation:
Formula: speed = distance / time
Values: distance = 50 m, time = 5 s
Calculation: speed = 50 m / 5 s = 10 m/s
Answer: The average speed of the cyclist is 10 m/s.
A remote controlled car moves at a constant speed. State the distance it will cover in 10 seconds if its speed is 2 m/s.
Formula: distance = speed × time
Values: speed = 2 m/s, time = 10 s
Calculation: distance = 2 m/s × 10 s = 20 m
Answer: The remote controlled car will cover a distance of 20 meters.
Define velocity.
Velocity is speed in a given direction. It is a vector quantity, meaning it has both magnitude (speed) and direction. (2 marks)
A remote control car travels at a constant speed of 0.5 m/s. State two different velocities the car could have.
The car could have a velocity of 0.5 m/s North and a velocity of 0.5 m/s South. Velocity includes both speed and direction so changing the direction changes the velocity, even if the speed remains constant.
A cyclist travels 180 meters in 30 seconds. Calculate the cyclist's average speed.
Average speed = total distance / total time
Average speed = 180 m / 30 s
Average speed = 6.0 m/s
Explanation: The average speed is found by dividing the total distance travelled by the total time taken.
State the equation used to calculate average speed. Define each term in the equation.
Average speed = total distance / total time
Total distance: The total length of the path travelled by an object (measured in meters, m).
Total time: The duration of the journey (measured in seconds, s).
Average speed: The rate at which an object covers distance (measured in meters per second, m/s).
A car travels 200m in 10 seconds, then remains stationary for 5 seconds, and finally travels another 100m in 5 seconds. Calculate the average speed for the entire journey.
Average speed = Total distance / Total time
Total distance = 200m + 0m + 100m = 300m
Total time = 10s + 5s + 5s = 20s
Average speed = 300m / 20s = 15 m/s
Explanation: Average speed considers the total distance traveled over the entire duration, including any periods of rest.
Describe how you would determine the speed of an object from a distance-time graph.
1. Identify two points on the graph.
2. Determine the change in distance (vertical axis) between these two points.
3. Determine the corresponding change in time (horizontal axis) between the same two points.
4. Divide the change in distance by the change in time. This gives the speed during that time interval.
Speed = (Change in distance) / (Change in time) which is the gradient of the graph.
The following data shows the distance travelled by a toy car at different times: Time (s) | Distance (m) ------- | -------- 0 | 0 1 | 0.5 2 | 2.0 3 | 4.5 4 | 8.0 Determine whether the car is accelerating, decelerating, moving at constant speed, or at rest. Explain your reasoning.
The car is accelerating.
The distance travelled increases more with each passing second. If the distance increased linearly with time, the speed would be constant. Here, the increase in distance each second increases, implying increasing speed.
Describe the motion of an object whose distance-time graph is a horizontal line.
The object is at rest.
Explanation: A horizontal line on a distance-time graph indicates that the distance from the starting point is not changing over time. This means the object is stationary and not moving.
A cyclist travels along a straight road. The distance-time graph shows that the cyclist travels 200m in 25 seconds at a constant speed. Calculate the cyclist's speed during this time.
Formula: speed = distance / time
Working: speed = 200m / 25s
Answer: speed = 8 m/s
Explanation: The gradient of a distance-time graph represents speed. Since the section is a straight line, the speed is constant and can be calculated by dividing the distance traveled by the time taken.
Explain why the slope of a straight-line section on a distance-time graph represents the object's speed.
The slope of a distance-time graph is calculated as the change in distance divided by the change in time (rise over run). This calculation directly corresponds to the definition of speed, which is the rate of change of distance with respect to time. Therefore, a steeper slope indicates a greater change in distance per unit time, meaning a higher speed. A straight line indicates that this rate of change (speed) is constant.
A cyclist accelerates uniformly from rest to a speed of 8 m/s in 5 seconds. Calculate the distance travelled by the cyclist during this acceleration.
Formula: Distance = Area under speed-time graph.
Since acceleration is constant, the graph is a straight line. The area is a triangle.
Area of triangle = 0.5 * base * height
Distance = 0.5 * 5 s * 8 m/s = 20 m
Answer: The distance travelled is 20 m.
A car moves at a constant speed of 15 m/s for 10 seconds, then decelerates uniformly to rest in 5 seconds. Sketch a speed-time graph for this motion, and then determine the total distance travelled by the car.
Speed-Time Graph: The graph consists of a horizontal line at 15 m/s for 10 seconds, followed by a straight line sloping downwards to 0 m/s at 15 seconds (10 + 5 = 15).
Total Distance:
Distance during constant speed = 15 m/s * 10 s = 150 m
Distance during deceleration = 0.5 * 5 s * 15 m/s = 37.5 m
Total Distance = 150 m + 37.5 m = 187.5 m
Answer: The total distance travelled is 187.5 m.
A tennis ball is dropped from rest. Calculate the speed of the ball after 1.5 seconds, assuming air resistance is negligible and the acceleration of free fall is 9.8 m/s².
Formula: v = u + at
Working: v = 0 + (9.8 m/s²)(1.5 s) = 14.7 m/s
Answer: 14.7 m/s
Explanation: We used the equation of motion v=u+at, where 'v' is final velocity, 'u' is initial velocity (0 in this case), 'a' is acceleration (9.8 m/s²), and 't' is time (1.5 s).
State the approximate value of the acceleration of free fall near the Earth's surface, including the appropriate unit.
Answer: Approximately 9.8 m/s²
Explanation: The acceleration due to gravity is a constant value near the Earth's surface, causing objects to accelerate downwards at a rate of approximately 9.8 meters per second squared.
A car accelerates from 10 m/s to 25 m/s in 5 seconds. Calculate the acceleration of the car.
Acceleration is the change in velocity divided by time.
Formula: a = (v - u) / t
Where:
a = acceleration
v = final velocity (25 m/s)
u = initial velocity (10 m/s)
t = time (5 s)
a = (25 - 10) / 5 = 15 / 5 = 3 m/s²
Answer: The acceleration of the car is 3 m/s².
State the effect on the acceleration of an object if the change in velocity is doubled but the time taken remains the same.
Acceleration is defined as the change in velocity per unit time. If the change in velocity is doubled while the time remains constant, the acceleration will also double.
Explanation: Since acceleration = (change in velocity) / time, doubling the numerator (change in velocity) while keeping the denominator (time) constant will double the result (acceleration).
A car accelerates from rest. Its speed is recorded at different times. At t = 2.0 s, the speed is 4.0 m/s. At t = 4.0 s, the speed is 8.0 m/s. At t = 6.0 s, the speed is 10.0 m/s. Determine whether the car is moving with constant acceleration, changing acceleration or neither between t=2.0s and t=6.0s. Show your working.
To determine if the acceleration is constant, calculate the acceleration between t=2.0s and t=4.0s, and then between t=4.0s and t=6.0s.
Acceleration (2-4s) = (8.0 m/s - 4.0 m/s) / (4.0 s - 2.0 s) = 4.0 m/s / 2.0 s = 2.0 m/s²
Acceleration (4-6s) = (10.0 m/s - 8.0 m/s) / (6.0 s - 4.0 s) = 2.0 m/s / 2.0 s = 1.0 m/s²
Since the acceleration is different over the two time intervals, the car is moving with *changing acceleration* between t=2.0s and t=6.0s.
Sketch a speed-time graph to represent the motion of an object that is moving with changing acceleration. Label the axes.
The graph should have time on the x-axis and speed on the y-axis. The line representing changing acceleration should be a curve, not a straight line.
The axes should be labelled clearly:
X-axis: Time (s)
Y-axis: Speed (m/s)
Explanation: A curved line on a speed-time graph indicates that the gradient (acceleration) is changing with time, representing changing acceleration.
A car accelerates from rest. A speed-time graph shows that its speed increases uniformly from 0 m/s to 25 m/s in 5.0 s. Calculate the acceleration of the car.
Formula: acceleration = (change in speed) / (time taken)
Working:
acceleration = (25 m/s - 0 m/s) / 5.0 s
acceleration = 5.0 m/s²
Explanation: The acceleration is calculated by finding the gradient of the speed-time graph during the period of uniform acceleration. The change in speed (rise) is divided by the change in time (run).
A cyclist is accelerating. Explain how you would determine the cyclist's acceleration from a speed-time graph.
To determine acceleration from a speed-time graph, first identify a straight section of the graph where the speed is changing uniformly. Then, calculate the gradient of this section. The gradient represents the acceleration. The gradient is calculated by dividing the change in speed (vertical change) by the change in time (horizontal change) for that section of the graph. Acceleration = Δspeed / Δtime
A car is travelling at 25 m/s when the driver applies the brakes. The car decelerates uniformly at 2.0 m/s². Calculate the distance the car travels before coming to a complete stop.
Formula: v² = u² + 2as
Where:
v = final velocity (0 m/s)
u = initial velocity (25 m/s)
a = acceleration (-2.0 m/s²)
s = distance (unknown)
Working:
0² = 25² + 2 * (-2.0) * s
0 = 625 - 4s
4s = 625
s = 625 / 4
s = 156.25 m
Answer: The car travels 156.25 m before stopping. The acceleration is negative (deceleration), hence the minus sign in the calculation.
Explain why a deceleration is considered a negative acceleration. Give an example of a situation where an object experiences deceleration.
Deceleration is a negative acceleration because it represents a decrease in velocity over time in the direction of motion. Acceleration is defined as the rate of change of velocity. If the velocity is decreasing, the change in velocity is negative, hence the acceleration is negative.
A skydiver of mass 75 kg jumps from a plane. Calculate the terminal velocity they reach if the air resistance force, *F*, is given by *F* = 0.6*v*², where *v* is the velocity in m/s. (Assume *g* = 9.8 m/s²).
At terminal velocity, the air resistance force equals the skydiver's weight. Therefore, 0.6*v*2 = *mg*.
Rearranging: *v*2 = *mg* / 0.6
*v*2 = (75 kg * 9.8 m/s2) / 0.6 = 1225 m2/s2
*v* = √1225 m2/s2 = 35 m/s
Answer: 35 m/s. The skydiver stops accelerating when the upward force of air resistance balances the downward force of gravity.
Describe and explain how the velocity of a stone changes from the moment it is dropped from a tall building, until it hits the ground, considering the effect of air resistance.
Initially, the stone accelerates downwards due to gravity. As its velocity increases, the air resistance opposing its motion also increases. The resultant force is the difference between the stone's weight and the air resistance. As the stone continues to accelerate, air resistance continues to increase. Eventually, the air resistance becomes equal in magnitude to the stone's weight. At this point, the resultant force is zero, and the stone stops accelerating, reaching its terminal velocity. The velocity remains constant until the stone hits the ground.
Key Questions: Motion
Define velocity.
Velocity is speed in a given direction. It is a vector quantity, meaning it has both magnitude (speed) and direction. (2 marks)
State the equation used to calculate average speed. Define each term in the equation.
Average speed = total distance / total time
Total distance: The total length of the path travelled by an object (measured in meters, m).
Total time: The duration of the journey (measured in seconds, s).
Average speed: The rate at which an object covers distance (measured in meters per second, m/s).
Tips to avoid common mistakes in Motion
- ● Carefully interpret speed-time graphs: constant non-zero speed is a horizontal line *above* the x-axis; a flat line *on* the x-axis means 'stopped'.
- ● When working with motion graphs, first highlight the exact time interval the question refers to, then decide if you're finding constant speed or an acceleration.
- ● Before calculating anything from a graph, take a moment to verify the scale on both axes.
- ● Before answering, circle the labels on the x-axis and y-axis to keep your interpretation accurate.
- ● Before analysing a motion graph, double-check what the x-axis and y-axis represent.
More topics in Unit 1 — Motion, forces and energy
Motion sits alongside these Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
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Cambridge syllabus keywords to use in your answers
These are the official Cambridge 0625 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Key terms covered in this Motion deck
Every term below is defined in the flashcards above. Use the list as a quick recall test before your exam — if you can't define one of these in your own words, flip back to that card.
Related Physics guides
Long-read articles that go beyond the deck — cover the whole subject's common mistakes, high-yield content and revision pacing.
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