IGCSE Physics | Practice Test — Answers & Worked Solutions
Write down the definition of speed.
Speed is the distance travelled per unit time.
Write down the definition of velocity and explain how it differs from speed.
Velocity is speed in a given direction.
Write down the equation for speed and state the unit of each quantity.
- $v$ = speed — unit: m/s (metres per second)
- $s$ = distance — unit: m (metres)
- $t$ = time — unit: s (seconds)
State the approximate value of $g$, the acceleration of free fall near the surface of the Earth. Include the unit.
On a distance–time graph, what does a horizontal (flat) line tell you about the motion of an object?
A horizontal line means the distance is not changing. The object is at rest (stationary).
On a speed–time graph, a straight line slopes downward from a positive value to zero. What is the object doing?
The object is decelerating — its speed is decreasing at a constant rate until it stops.
What physical quantity is represented by the area under a speed–time graph?
Write down the definition of acceleration, state the equation used to calculate it, and give the unit of acceleration. Supplement
Acceleration is the change in velocity per unit time.
$$a = \frac{\Delta v}{\Delta t}$$- $a$ = acceleration — unit: m/s²
- $\Delta v$ = change in velocity (final velocity − initial velocity) — unit: m/s
- $\Delta t$ = time taken — unit: s
Define terminal velocity. State the condition (in terms of forces) that causes a falling object to reach terminal velocity. Supplement
Terminal velocity is the maximum constant speed that a falling object reaches when the forces acting on it are balanced.
On a speed–time graph, how can you tell the difference between constant acceleration and changing (non-constant) acceleration? Supplement
- Constant acceleration → the speed–time graph shows a straight sloping line (the gradient is constant).
- Changing acceleration → the speed–time graph shows a curved line (the gradient is changing).
A car travels 200 m in 8 s. Calculate its speed.
- Formula: $$v = \frac{s}{t}$$
- Given: $s = 200\ \text{m}$, $t = 8\ \text{s}$
- Substitute: $$v = \frac{200}{8}$$
- Calculate: $$v = 25\ \text{m/s}$$
A cyclist moves at a constant speed of 6 m/s. Calculate the distance the cyclist travels in 45 s.
- Formula: $s = v \times t$ (rearranged from $v = s/t$)
- Given: $v = 6\ \text{m/s}$, $t = 45\ \text{s}$
- Substitute: $$s = 6 \times 45$$
- Calculate: $$s = 270\ \text{m}$$
A ball rolls at a speed of 3 m/s and covers a distance of 27 m. Calculate the time taken.
- Formula: $t = s \div v$ (rearranged from $v = s/t$)
- Given: $v = 3\ \text{m/s}$, $s = 27\ \text{m}$
- Substitute: $$t = \frac{27}{3}$$
- Calculate: $$t = 9\ \text{s}$$
A bus travels 12 km in 20 minutes. Calculate the average speed of the bus in m/s.
- Convert units to metres and seconds:
Distance: $12\ \text{km} = 12 \times 1000 = 12\,000\ \text{m}$
Time: $20\ \text{min} = 20 \times 60 = 1\,200\ \text{s}$ - Formula: $$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$$
- Substitute: $$\text{average speed} = \frac{12\,000}{1\,200}$$
- Calculate: $$\text{average speed} = 10\ \text{m/s}$$
A distance–time graph shows a straight line from the point (0 s, 0 m) to the point (5 s, 30 m). Calculate the speed of the object.
- Speed = gradient of the distance–time graph
- Formula: $$\text{speed} = \frac{\Delta s}{\Delta t}$$
- $\Delta s = 30 – 0 = 30\ \text{m}$, $\Delta t = 5 – 0 = 5\ \text{s}$
- Substitute: $$\text{speed} = \frac{30}{5}$$
- Calculate: $$\text{speed} = 6\ \text{m/s}$$
A speed–time graph shows a horizontal line at $v = 8$ m/s from $t = 0$ s to $t = 10$ s. Calculate the distance travelled by the object.
- Distance = area under the speed–time graph
- Shape: a rectangle (constant speed)
- Area = width × height = time × speed $$\text{Distance} = 10\ \text{s} \times 8\ \text{m/s}$$
- Calculate: $$\text{Distance} = 80\ \text{m}$$
A speed–time graph shows a straight line from (0 s, 0 m/s) to (6 s, 12 m/s). Calculate the distance travelled by the object.
- Distance = area under the speed–time graph
- Shape: a triangle (speed starts at zero and increases)
- Area = ½ × base × height $$\text{Distance} = \frac{1}{2} \times 6\ \text{s} \times 12\ \text{m/s}$$
- Calculate: $$\text{Distance} = 36\ \text{m}$$
A motorcycle accelerates from 10 m/s to 30 m/s in 5 s. Calculate the acceleration. Supplement
- Formula: $$a = \frac{\Delta v}{\Delta t}$$
- Given: initial velocity = 10 m/s, final velocity = 30 m/s, $\Delta t = 5\ \text{s}$
- Change in velocity: $\Delta v = 30 – 10 = 20\ \text{m/s}$
- Substitute: $$a = \frac{20}{5}$$
- Calculate: $$a = 4\ \text{m/s}^2$$
A train decelerates from 40 m/s to 10 m/s in 6 s. Calculate the acceleration. State whether your answer is positive or negative and explain what the sign tells you. Supplement
- Formula: $$a = \frac{\Delta v}{\Delta t}$$
- Given: initial velocity = 40 m/s, final velocity = 10 m/s, $\Delta t = 6\ \text{s}$
- Change in velocity: $\Delta v = 10 – 40 = -30\ \text{m/s}$
- Substitute: $$a = \frac{-30}{6}$$
- Calculate: $$a = -5\ \text{m/s}^2$$
The answer is negative. This shows the train is decelerating — its velocity is decreasing. A negative acceleration means the acceleration acts in the opposite direction to the motion.
A speed–time graph shows a straight line from (0 s, 4 m/s) to (10 s, 14 m/s). Calculate the distance travelled during this time.
- Distance = area under the speed–time graph
- Shape: a trapezium (speed does not start from zero)
- Formula for trapezium area: $$\text{Distance} = \frac{1}{2}(v_1 + v_2) \times t$$
- Given: $v_1 = 4\ \text{m/s}$, $v_2 = 14\ \text{m/s}$, $t = 10\ \text{s}$
- Substitute: $$\text{Distance} = \frac{1}{2}(4 + 14) \times 10 = \frac{1}{2} \times 18 \times 10$$
- Calculate: $$\text{Distance} = 90\ \text{m}$$
A speed–time graph for a car journey shows three phases:
- Phase 1 — from $t = 0$ s to $t = 5$ s: speed increases from 0 to 20 m/s (straight line)
- Phase 2 — from $t = 5$ s to $t = 15$ s: speed is constant at 20 m/s
- Phase 3 — from $t = 15$ s to $t = 20$ s: speed decreases from 20 m/s to 0 (straight line)
(a) Describe the motion. (b) Calculate the acceleration in Phase 1. (c) Calculate the total distance.
Speed–time graph for the car journey (shaded areas represent distance travelled in each phase)
- Phase 1 (0–5 s): The car is accelerating — speed increases at a constant rate from 0 to 20 m/s.
- Phase 2 (5–15 s): The car moves at constant speed — 20 m/s with no acceleration.
- Phase 3 (15–20 s): The car is decelerating — speed decreases at a constant rate from 20 m/s back to 0.
- Formula: $$a = \frac{\Delta v}{\Delta t}$$
- $\Delta v = 20 – 0 = 20\ \text{m/s}$, $\Delta t = 5 – 0 = 5\ \text{s}$
- Substitute: $$a = \frac{20}{5}$$
- $$a = 4\ \text{m/s}^2$$
- Distance = total area under the speed–time graph
- Phase 1 area (triangle): $$d_1 = \frac{1}{2} \times 5 \times 20 = 50\ \text{m}$$
- Phase 2 area (rectangle): $$d_2 = 10 \times 20 = 200\ \text{m}$$
- Phase 3 area (triangle): $$d_3 = \frac{1}{2} \times 5 \times 20 = 50\ \text{m}$$
- Total: $$d = 50 + 200 + 50 = 300\ \text{m}$$
A speed–time graph for an accelerating object shows a straight line from the point (2 s, 6 m/s) to the point (8 s, 18 m/s). Calculate the acceleration of the object. Supplement
- Acceleration = gradient of the speed–time graph
- Formula: $$a = \frac{\Delta v}{\Delta t}$$
- $\Delta v = 18 – 6 = 12\ \text{m/s}$, $\Delta t = 8 – 2 = 6\ \text{s}$
- Substitute: $$a = \frac{12}{6}$$
- Calculate: $$a = 2\ \text{m/s}^2$$
An object is released from rest and falls freely under gravity with no air resistance. (a) State the acceleration. (b) Calculate the speed after 4 s. (c) Sketch the speed–time graph.
The acceleration is equal to $g$, the acceleration of free fall, which is approximately constant at 9.8 m/s² near Earth’s surface.
- Use: $$a = \frac{\Delta v}{\Delta t} \implies \Delta v = a \times \Delta t$$
- Object released from rest: initial speed = 0 m/s
- $\Delta v = 9.8 \times 4$
Written method: $(10 – 0.2) \times 4 = 40 – 0.8 = 39.2\ \text{m/s}$ - Final speed = initial speed + $\Delta v = 0 + 39.2 = 39.2\ \text{m/s}$
Speed–time graph for free fall (no air resistance). The line is straight because the acceleration is constant at 9.8 m/s².
A stone is dropped from rest through the air. Air resistance acts on the stone as it falls. (a) State the two forces. (b) Explain why acceleration decreases. (c) Describe terminal velocity and the force relationship. Supplement
2. Air resistance (drag) — acting upward, opposing the motion
- At the start: weight is much greater than air resistance → large net downward force → large downward acceleration.
- As the stone speeds up, air resistance increases (drag increases with speed).
- The net force (weight − air resistance) therefore decreases.
- A smaller net force means a smaller acceleration — the stone is still speeding up, but more slowly.
At terminal velocity, the stone is falling at a constant speed — there is no acceleration. This happens because:
The stone has reached its maximum speed and will continue to fall at this speed unless something changes.
A parachutist jumps from an aircraft and falls through the air. (a) Explain initial rapid acceleration. (b) Explain how air resistance affects acceleration. (c) State the condition for terminal velocity. (d) Describe what happens when the parachute opens. Supplement
At the moment of jumping, speed is zero, so air resistance is very small. The weight of the parachutist is much greater than the air resistance. There is a large net downward force, which causes a large downward acceleration.
As the parachutist falls faster:
- Air resistance increases (drag increases with speed).
- The net downward force (weight − air resistance) decreases.
- The acceleration decreases — the parachutist is still speeding up, but more slowly.
The net force is zero, so acceleration = 0 and speed is constant.
When the parachute opens, the air resistance suddenly becomes much greater than the weight. This means:
- The net force is now upward (air resistance > weight).
- This upward net force causes an upward acceleration — a negative acceleration (deceleration).
- The parachutist’s speed decreases rapidly.
- As speed decreases, air resistance decreases until a new (much lower) terminal velocity is reached.
