These are real past-paper questions. Think of your answer first, then open the explanation.
Oct/Nov 2024 · Paper 1 · Kinematics
A student takes 20 minutes to walk from X to Y at a speed of 3.0 km / h. She turns 90° at Y and walks at a speed of 4.0 km / h for 15 minutes to reach Z.
What is her average speed from X to Z?
- 2.4 km / h
- 3.4 km / h
- 3.5 km / h
- 5.0 km / h
Show answer and explanation
Answer: B. 3.4 km / h
Distance XY = 3.0 × (20/60) = 1.0 km; distance YZ = 4.0 × (15/60) = 1.0 km; total distance = 2.0 km; total time = 35 min = 7/12 h; average speed = 2.0 ÷ (7/12) ≈ 3.4 km/h.
May/June 2022 · Paper 1 · Kinematics
A driver drives a car at uniform velocity along a road. The driver sees a hazard and applies the brakes.
What is the car's motion as it moves through the thinking distance and as it moves through the braking distance?
- thinking distance: deceleration; braking distance: deceleration
- thinking distance: deceleration; braking distance: uniform velocity
- thinking distance: uniform velocity; braking distance: deceleration
- thinking distance: uniform velocity; braking distance: uniform velocity
Show answer and explanation
Answer: C. thinking distance: uniform velocity; braking distance: deceleration
During the thinking distance the driver has not yet applied the brakes, so the car still moves at uniform velocity; once the brakes are applied during the braking distance the car decelerates.
Oct/Nov 2022 · Paper 1 · Kinematics
A driver sees a hazard ahead. The car travels a thinking distance while the driver reacts and a braking distance while the brakes act.
Which expression is equal to the stopping distance?
- braking distance + thinking distance
- braking distance – thinking distance
- braking distance × thinking distance
- braking distance ÷ thinking distance
Show answer and explanation
Answer: A. braking distance + thinking distance
Stopping distance is the total distance covered from seeing the hazard to stopping, which is the thinking distance plus the braking distance.
May/June 2024 · Paper 2 · Kinematics
The speeds of the ball at A, B, C and D are vA, vB, vC and vD respectively. Arrange these four speeds in order from slowest to fastest.
Show model answer
vA, vC, vD, vB (slowest to fastest)
The ball's speed depends on how far it has fallen from its release height at A (energy conservation, smooth track, negligible air resistance): vA = 0 (released from rest, so slowest); C is a hump close in height to A, so speed there is still low; D is on the lower flat section, so faster than C; B is at the bottom of the deepest dip, the lowest point, so the ball is fastest there.