CAIE IGCSE Physics Motion Question Bank
Practise speed, velocity and acceleration calculations, interpreting distance-time and speed-time graphs and explaining free fall, drag and terminal velocity.
- Syllabus
- 2026–2028
- Course
- Physics 0625
Practise speed, velocity and acceleration calculations, interpreting distance-time and speed-time graphs and explaining free fall, drag and terminal velocity.
Fig. 3.1 shows a model of a wind turbine used to demonstrate the use of wind energy to generate electricity. The wind is blowing towards the model, as shown.

Fig. 3.1
The mass of air passing through the circular area swept out by the turbine blades each second is 7.5 kg . The kinetic energy of the air that passes through this circular area each second is 240 J .
Calculate the speed of the air.
KE=21mv2 in any form ORv2=2×KE/m
OR 240=21×7.5v2
C1
v2=2×240/7.5 OR (v=){2×240/7.5}OR(v=)↓{2KE/m}
C1
=8.0 m/s
A1
Fig. 2.1 shows a motorcyclist accelerating along a straight horizontal section of track.

Fig. 2.1
The motorcyclist and motorcycle have a combined mass of 240 kg .
On the straight horizontal section of the track, the motorcyclist accelerates from rest at 7.2 m/s2.
The motorcyclist reaches the end of the straight section of track in 5.3 s .
Calculate the speed of the motorcyclist at the end of the straight section.
speed =
(speed =)38 m/s
A2
a=Δv/ΔtOR(Δv=)aΔtOR(Δv=)7.2×5.3
C1
At the end of the straight section, the track remains horizontal but bends to the right, as shown in Fig. 2.1.
When the motorcyclist reaches the bend, she travels around the bend in a circular path at a constant speed.
Describe what happens to the velocity of the motorcyclist as she travels around the bend at constant speed.
(velocity) changes (as direction of motion changes) OR direction (of velocity) changes
B1
Fig. 1.1 is the distance-time graph for a moving car.

Fig. 1.1
On Fig. 1.1, mark a point P where the acceleration of the car is zero.
P marked on line between t=0 s and t=30 s
B1
Determine:
the speed of the car at time t=15 s
speed = [2]
( v= ) gradient or 150 / 30 or appropriate division using other points
C1
5.0 m/s
A1
the average speed of the car between time t=30 s and time t=45 s.
average speed = [2]
(v=) x / t or (300-150) /(45-30) or 150 / 15
C1
10 m/s
A1
At time t=45 s, the car starts to decelerate. At time t=55 s and at a distance of 400 m from the starting point, the car stops. It then remains stationary for 5.0 s .
On Fig. 1.1, draw a possible continuation of the distance-time graph.
[Total: 8]
gradient decreasing
B1
smooth transition to horizontal and line not too thick
B1
horizontal to ( 60 s,400 m )
B1