Free Motion questions for IGCSE Physics (0625), part of Motion, forces & energy. Each question has the answer and the marking points an examiner would look for.
Generate a full Motion worksheet →All Motion, forces & energy questionsA skydiver is falling at terminal velocity. She then opens her parachute.
Which row describes the skydiver immediately after the parachute opens?
| direction of motion | direction of resultant force | speed | |
|---|---|---|---|
| A | downwards | upwards | decreasing |
| B | downwards | downwards | decreasing |
| C | downwards | upwards | increasing |
| D | upwards | upwards | decreasing |
Answer: A
Which quantity is a vector?
Answer: A
A skydiver jumps from a plane. After some time she reaches a constant speed called terminal velocity.
Describe how her acceleration changes from the moment she jumps until she reaches terminal velocity.
Explain why she reaches a terminal velocity.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | starts at g, decreases to zero acceleration is 9.8 m/s² at first B1 it decreases to zero as air resistance increases B1 | 2 |
| (b) | air resistance increases until equal to weight air resistance increases with speed B1 when it equals her weight the resultant force is zero B1 | 2 |
A car of mass 1500 kg travels along a straight, horizontal road at 24 m/s. The driver sees a hazard. After a delay of 0.8 s the brakes are applied and the car decelerates uniformly at its maximum deceleration of 6 m/s² until it stops.
Sketch the speed–time graph for the car from the moment the driver sees the hazard until it stops. Label the values on both axes.
Determine the total distance travelled from the moment the driver sees the hazard until the car stops.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | horizontal line at 24 m/s to 0.8 s; straight line down to 0 at 4.80 s horizontal line at 24 m/s for 0.8 s B1 straight line sloping down to zero B1 reaching zero at 4.80 s B1 | 3 |
| (b) | 67.2 m distance at constant speed = 24 × 0.8 = 19.2 m C1 distance while decelerating = area = ½ × 24 × 4.00 = 48.0 m C1 67.2 m A1 | 3 |
A distance–time graph shows a cyclist’s journey.
State what a horizontal line on a distance–time graph shows.
State how the speed can be found from a distance–time graph.
Describe how a curved line that becomes steeper should be interpreted.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | stationary the cyclist is stationary (not moving) B1 | 1 |
| (b) | gradient the gradient of the graph B1 | 1 |
| (c) | accelerating the cyclist is accelerating (speed increasing) B1 | 1 |
Fig. 6.1 shows the speed–time graph for a car of mass 800 kg. The car accelerates uniformly, then the driver brakes and the car decelerates uniformly to rest.
State what is meant by acceleration.
Calculate the acceleration of the car during the first part of the journey.
Calculate the resultant force on the car while it accelerates.
Calculate the total distance travelled from the start of the acceleration until the car stops.
Explain why the braking force must be larger if the car carries passengers and luggage, to stop in the same time.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | rate of change of velocity change in velocity per unit time B1 | 1 |
| (b) | 1.20 m/s² (18 − 12)/5 C1 1.20 m/s² A1 | 2 |
| (c) | 960 N F = ma = 800 × 1.20 C1 960 N A1 | 2 |
| (d) | 111 m area under graph: 75 m while accelerating C1 36 m while braking C1 111 m (accept 111 m) A1 | 3 |
| (e) | greater mass needs greater force for the same deceleration the mass is larger B1 F = ma so a larger force is needed for the same deceleration B1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in