Cambridge IGCSE Physics 0625 – Topic 1.2: Motion
Objective
Define speed, velocity and average speed; use the equations (1.2.2 v = s/t) and (1.2.4 \(\bar v\) = total distance / total time); sketch and interpret distance‑time and speed‑time graphs; calculate speed from the gradient of a distance‑time graph and distance from the area under a speed‑time graph; state the free‑fall acceleration (1.2.8 g ≈ 9.8 m s⁻²).
1. Speed (scalar) – 1.2.1
Definition: Speed is the total distance travelled divided by the time taken. It has magnitude only (no direction).
\( v = \dfrac{s}{t} \) (1.2.2)
- v – speed (SI unit: m s⁻¹, common unit: km h⁻¹)
- s – distance travelled (m or km)
- t – time taken (s or h)
Units & conversions
- SI unit: metres per second (m s⁻¹).
- Everyday unit: kilometres per hour (km h⁻¹).
- Conversion factors:
1 m s⁻¹ = 3.6 km h⁻¹ and 1 km h⁻¹ = 0.278 m s⁻¹.
Rearranging the formula
- Distance:
s = v × t
- Time:
t = s / v
2. Velocity (vector) – 1.2.3
Velocity has both magnitude and direction. It is written with an arrow or bold type, e.g. →v or v. The magnitude of velocity is the speed.
Diagram example: a car moving east at 20 m s⁻¹ – the arrow points east, its length proportional to 20 m s⁻¹.
3. Average speed – 1.2.4
Average speed = total distance travelled ÷ total time taken. It does not require uniform motion.
\( \bar{v} = \dfrac{\text{total distance}}{\text{total time}} \)
4. Graphical interpretation – 1.2.5
Graph‑interpretation rules (syllabus wording)
| Graph type | Feature | What it means (syllabus code) |
|---|
| Distance‑time | Horizontal segment | Object at rest (1.2.5a) |
| Distance‑time | Straight line, constant gradient | Uniform speed; gradient = speed (1.2.6) |
| Distance‑time | Steeper gradient | Higher speed (1.2.6) |
| Speed‑time | Horizontal line | Constant speed (1.2.5b) |
| Speed‑time | Straight line, constant gradient | Uniform acceleration; gradient = acceleration (1.2.7) |
| Speed‑time | Area under the curve | Distance travelled (1.2.7) |
4.1 Distance‑time graphs
- Gradient = speed (m s⁻¹ or km h⁻¹).
- Horizontal segment = object at rest.
- Steeper gradient = higher speed.
4.2 Speed‑time graphs
- Gradient = acceleration (m s⁻²).
- Area under the graph = distance travelled.
- Horizontal line = constant speed (uniform motion).
Quick sketch exercise
Draw a distance‑time graph for a car that:
- Travels at 10 m s⁻¹ for 5 s,
- Stops for 2 s,
- Accelerates uniformly to 20 m s⁻¹ in 3 s.
5. Acceleration – 1.2.7
Acceleration is the rate of change of velocity.
\( a = \dfrac{\Delta v}{\Delta t} \)
- a – acceleration (m s⁻²)
- Δv – change in velocity (m s⁻¹)
- Δt – time interval (s)
6. Free‑fall acceleration – 1.2.8
Near the Earth’s surface, all objects accelerate downwards at
\( g \approx 9.8\;\text{m s}^{-2} \)
Direction is downwards; the vector is written →g.
7. Measurement techniques & experimental skills
- Distance: ruler or measuring tape; record to the nearest mm (or cm).
- Time: digital stopwatch; record to the nearest 0.01 s.
- Take ≥ 3 trials, calculate the mean.
- Estimate uncertainties:
Δs ≈ ±0.5 cm (ruler) and Δt ≈ ±0.01 s (stopwatch).
Propagate to speed using \( \displaystyle \frac{Δv}{v} \approx \frac{Δs}{s} + \frac{Δt}{t} \).
- Present data in a tidy table, include units, and label graphs clearly.
Practical tip
When measuring the period of a pendulum, time 20 complete oscillations and divide by 20 – this reduces random error.
8. Scalar vs. vector notation
| Quantity | Scalar notation | Vector notation |
|---|
| Speed | v | — |
| Velocity | — | →v or v |
| Acceleration | — | →a or a |
9. Worked examples
Example 1 – Mixed units
A car travels 150 km in 2 h 15 min. Find its speed in (a) km h⁻¹ and (b) m s⁻¹.
- Convert time: 2 h 15 min = 2.25 h = 8100 s.
- Speed (km h⁻¹): \( v = \dfrac{150\;\text{km}}{2.25\;\text{h}} = 66.7\;\text{km h}^{-1} \).
- Speed (m s⁻¹): \( 66.7\;\text{km h}^{-1} \times 0.278 = 18.5\;\text{m s}^{-1} \).
Example 2 – Interpreting a speed‑time graph
The graph rises uniformly from 0 to 12 m s⁻¹ over 4 s, then stays constant for 6 s. Determine (a) the acceleration during the first 4 s, (b) the total distance travelled in the 10 s.
- Acceleration: gradient = Δv/Δt = 12 / 4 = 3 m s⁻².
- Distance = area under the graph = (½ × 12 × 4) + (12 × 6) = 24 + 72 = 96 m.
Example 3 – Average speed (non‑uniform motion)
A hiker walks 8 km to a hilltop in 2 h, rests 0.5 h, then walks back 8 km in 2 h. Find the average speed for the whole journey.
- Total distance = 8 km + 8 km = 16 km.
- Total time = 2 h + 0.5 h + 2 h = 4.5 h.
- \( \bar{v} = \dfrac{16\;\text{km}}{4.5\;\text{h}} = 3.56\;\text{km h}^{-1} \) (≈ 3.6 km h⁻¹).
Example 4 – Uncertainty calculation
Distance measured = 5.00 m ± 0.01 m, time measured = 2.00 s ± 0.02 s. Find speed and its percentage uncertainty.
- Speed: \( v = 5.00/2.00 = 2.50\;\text{m s}^{-1} \).
- Relative uncertainties: Δs/s = 0.01/5.00 = 0.2 %; Δt/t = 0.02/2.00 = 1.0 %.
- Combined: Δv/v ≈ 0.2 % + 1.0 % = 1.2 % → Δv ≈ 0.012 × 2.50 ≈ 0.03 m s⁻¹.
- Result: \( v = 2.50 ± 0.03\;\text{m s}^{-1} \).
10. Common mistakes
- Confusing speed (scalar) with velocity (vector).
- Using inconsistent units – always convert to SI before applying formulas.
- Omitting direction when dealing with velocity or acceleration.
- Forgetting the area‑under‑curve method for distance on speed‑time graphs.
- Neglecting multiple trials and the calculation of a mean value.
11. Quick‑check questions
- A car travels 120 km in 2 h. Give its speed in (a) km h⁻¹ and (b) m s⁻¹.
- If a runner’s speed is 5 m s⁻¹, how far will they run in 8 s?
- A train covers 300 m in 15 s. State its speed and say whether it is faster or slower than 10 m s⁻¹.
- Sketch a distance‑time graph for a person who walks 4 m s⁻¹ for 3 s, rests 2 s, then walks back 4 m s⁻¹ for 3 s.
- Calculate the acceleration of a bike that increases its speed from 2 m s⁻¹ to 14 m s⁻¹ in 4 s.
12. Summary table
| Quantity |
Symbol |
Formula (syllabus code) |
SI unit |
Common unit |
| Speed |
v |
v = s / t (1.2.2) |
m s⁻¹ |
km h⁻¹ |
| Velocity |
→v or v |
→v = s / t (with direction) (1.2.3) |
m s⁻¹ |
km h⁻¹ |
| Average speed |
\(\bar{v}\) |
\(\bar{v} = \dfrac{\text{total distance}}{\text{total time}}\) (1.2.4) |
m s⁻¹ |
km h⁻¹ |
| Acceleration |
a |
a = Δv / Δt (1.2.7) |
m s⁻² |
– |
| Free‑fall g |
g |
≈ 9.8 m s⁻² (1.2.8) |
m s⁻² |
– |
| Distance |
s |
s = v × t |
m |
km |
| Time |
t |
t = s / v |
s |
h |
13. Key points to remember
- Speed = distance ÷ time (scalar). Velocity = speed with a direction (vector).
- Always keep units consistent – convert to SI before using any formula.
- Average speed uses total distance and total time, not the arithmetic mean of instantaneous speeds.
- Gradient of a distance‑time graph = speed; gradient of a speed‑time graph = acceleration; area under a speed‑time graph = distance.
- Free‑fall acceleration is constant at ≈ 9.8 m s⁻² downwards.
- Record several measurements, calculate a mean, and estimate uncertainties.
- Convert between m s⁻¹ and km h⁻¹ using the factor 3.6 (or 0.278).