# Movement and Position

> Edexcel International GCSE Physics · 4PH1 (2017 spec)
> Source: https://www.owlsprep.com/study/edexcel-igcse-physics-s1-movement-and-position/

This guide covers all core content for Movement and Position in Edexcel IGCSE Physics (4PH1), including motion graphs, required formulas, practical skills and exam-standard calculation practice for both Paper 1 and Paper 2.

**Prerequisites:** [Basic SI units and graph interpretation](https://www.owlsprep.com/study/edexcel-igcse-physics-s1-intro-units-graphs/)

## Learning objectives

- Use correct SI units for all motion quantities in calculations and graph labelling
- Calculate average speed, acceleration and unknown motion values using the 3 required memorized formulas
- Plot, interpret and extract data from distance-time and velocity-time graphs
- Calculate acceleration from velocity-time graph gradients and distance from the area under velocity-time graphs
- Plan and carry out practical investigations of everyday object motion with valid control variables
- Apply consistent sign conventions for velocity, acceleration and deceleration correctly in calculations

## SI Units for Motion and Average Speed Calculations

All motion calculations in your exam require consistent SI units. You must memorize the correct units for all quantities used in this topic, as marks are awarded for correct unit quotation alongside numerical answers.

**Average speed** — Total distance travelled divided by total time taken

*Example:* A car that travels 120 m in 10 s has an average speed of 12 m/s

$$v_{avg} = \frac{\text{distance moved}}{\text{time taken}} = \frac{s}{t}$$

> **tip**
>
> To convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6. This is a common conversion required in exam questions.

**Worked example:** A cyclist completes a 5.4 km journey in 15 minutes. Calculate her average speed in m/s.

1. Convert all values to SI units: distance = 5.4 km = 5400 m, time = 15 minutes = 15 × 60 = 900 s.
2. $$v_{avg} = \frac{5400}{900} = 6$$
3. Final answer: 6 m/s, with correct units.

> **Exam tip:** Always show unit conversion steps in your working: examiners award marks for correct conversion even if your final numerical answer is wrong.

*Calculator:* allowed

## Plotting and Interpreting Distance-Time Graphs

Distance-time graphs show the total distance an object has travelled from a starting point against time taken. You will be asked to plot these from experimental data, extract data, and calculate speed from their gradient.

**Gradient of distance-time graph** — Equal to the speed of the object. A horizontal line means the object is stationary (speed = 0), a straight sloped line means constant speed.

**Worked example:** A distance-time graph for a walking student has a straight sloped section between t=0 s and t=20 s, where distance increases from 0 m to 30 m. Calculate the student's constant speed in this section.

1. Calculate the gradient using rise over run: change in y (distance) = 30 m - 0 m = 30 m, change in x (time) = 20 s - 0 s = 20 s.
2. $$v = \frac{30}{20} = 1.5$$
3. Final answer: 1.5 m/s.

**Check your understanding**

1. What does a horizontal line on a distance-time graph indicate?

   - Constant speed
   - Stationary object
   - Acceleration
   - Deceleration

   *Answer:* Stationary object

   *Why:* Correct: a horizontal line means distance is not changing over time, so the object is not moving.

> **Exam tip:** When drawing a distance-time graph from experimental data, always use a sharp pencil to plot points and draw a best-fit line, not dot-to-dot.

*Calculator:* allowed

## Acceleration Calculations and Velocity-Time Graphs

Acceleration is the rate of change of velocity, and velocity is a vector quantity so it accounts for direction of travel. Negative acceleration is called deceleration, and describes an object slowing down in the positive direction.

**Acceleration** — Change in velocity divided by time taken, units m/s²

*Notation:* $a = \frac{v - u}{t}$ where u = initial velocity, v = final velocity, t = time

> **note**
>
> Velocity can be negative if the object is travelling in the opposite direction to the chosen positive direction. Always use consistent sign conventions in your calculations.

**Worked example:** A car accelerates from rest to 22 m/s in 8 s. Calculate its acceleration.

1. Identify known values: u = 0 m/s (starts from rest), v = 22 m/s, t = 8 s.
2. $$a = \frac{22 - 0}{8} = 2.75$$
3. Final answer: 2.8 m/s² (2 significant figures, matching the data provided).

Velocity-time graphs plot velocity on the y-axis and time on the x-axis. The gradient of a velocity-time graph equals acceleration, a horizontal line means constant velocity (zero acceleration), a positive slope means acceleration, a negative slope means deceleration.

> **Exam tip:** Remember: speed is scalar, velocity is vector. If a question asks for velocity, you must include direction if required, and use correct sign conventions.

*Calculator:* allowed

## Calculating Acceleration and Distance from Velocity-Time Graphs

You are required to calculate two key values from velocity-time graphs: acceleration from the gradient, and total distance travelled from the area under the graph line (between the line and the time axis).

- To find gradient: draw a large right triangle on the sloped section, divide change in velocity (y-axis) by change in time (x-axis)
- To find area: split the space under the line into rectangles and triangles, calculate the area of each shape and add them together. For curved lines, count squares and estimate partial squares for the total area.

**Worked example:** A velocity-time graph for a bus has a sloped section from (0,0) to (5,10), then a horizontal section from (5,10) to (15,10), then a sloped section from (15,10) to (20,0). Calculate (a) the acceleration of the bus in the first 5 s, (b) the total distance travelled over the full 20 s.

1. Part a: Calculate gradient of first section: change in v = 10 - 0 = 10 m/s, change in t = 5 - 0 = 5 s.
2. $$a = \frac{10}{5} = 2 \text{ m/s}^2$$
3. Part b: Split area into 3 shapes: triangle (first 5s), rectangle (5-15s), triangle (15-20s).
4. $$Area_1 = 0.5 \times 5 \times 10 = 25 \text{ m}, Area_2 = 10 \times 10 = 100 \text{ m}, Area_3 = 0.5 \times 5 \times 10 = 25 \text{ m} \\ \text{Total distance} = 25 + 100 + 25 = 150 \text{ m}$$
5. Final answers: (a) 2 m/s², (b) 150 m.

> **Exam tip:** Always draw your gradient triangle as large as possible on the graph to reduce calculation error: marks are awarded for correct triangle construction even if your final value is slightly off.

*Calculator:* allowed

## $v^2 = u^2 + 2as$ Formula and Motion Practical Investigations

The third formula you must memorize is used when time taken is not given in the question, to relate initial speed, final speed, acceleration and distance travelled.

$$v^2 = u^2 + 2as$$

**Worked example:** A car travelling at 14 m/s applies brakes and decelerates at 7 m/s² until it stops. Calculate the distance it travels while braking.

1. Identify known values: u = 14 m/s, v = 0 m/s (stops), a = -7 m/s² (deceleration, so negative). We need to find s.
2. $$0^2 = 14^2 + 2(-7)s \\ 0 = 196 - 14s \\ 14s = 196 \\ s = 14$$
3. Final answer: 14 m.

You will also be examined on practical investigations of everyday object motion, for example measuring the speed of a toy car rolling down a ramp.

- Independent variable: e.g. height of ramp, dependent variable: speed of car, control variables: mass of car, surface of ramp
- Measure distance with a ruler or tape measure, measure time with a stopwatch or light gates for more accurate results
- Repeat measurements 3 times and calculate a mean to reduce random error

> **Exam tip:** Always state that light gates are more accurate than stopwatches for timing, as they eliminate human reaction time error, a common exam question point.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to convert units (e.g. using km instead of m, minutes instead of s) in calculations.
  - Why it fails: All motion formulas require SI units, so incorrect units lead to wrong numerical answers and lost marks.
  - Correct: Convert all values to m, s, m/s, m/s² before substituting into any formula, and show conversion steps clearly.
- **Wrong:** Using the gradient of a velocity-time graph to calculate distance, or area under a distance-time graph for speed.
  - Why it fails: You have mixed up the rules for the two graph types, leading to incorrect values.
  - Correct: Memorize: d-t graph gradient = speed, v-t graph gradient = acceleration, v-t graph area = distance.
- **Wrong:** Using a positive value for acceleration when an object is decelerating.
  - Why it fails: Deceleration is negative acceleration, so substituting a positive value will give incorrect results (e.g. for braking distance calculations).
  - Correct: Assign a consistent sign convention (e.g. forward = positive) so deceleration is always negative in calculations.
- **Wrong:** Rounding intermediate calculation steps to 1 or 2 significant figures.
  - Why it fails: Rounding early introduces rounding error, leading to a final answer that is outside the acceptable range for exam marks.
  - Correct: Keep 3-4 significant figures for all intermediate steps, only round the final answer to match the least number of sig figs in the question data.
- **Wrong:** Drawing dot-to-dot lines instead of best-fit lines for experimental graph data.
  - Why it fails: Dot-to-dot lines do not account for random error in measurements, and will lose you marks for graph plotting.
  - Correct: Use a ruler to draw a straight or curved best-fit line that passes as close to as many plotted points as possible, ignoring obvious anomalies.

## Cheatsheet

| Quantity | Formula | Graph Rule | Units |
| --- | --- | --- | --- |
| Average speed | $v_{avg} = \frac{s}{t}$ | Gradient of d-t graph | m/s |
| Acceleration | $a = \frac{v-u}{t}$ | Gradient of v-t graph | m/s² |
| Distance travelled | $s = v_{avg}t$ | Area under v-t graph | m |
| No time given | $v^2 = u^2 + 2as$ | - | All SI units |
| Unit conversion | km/h → m/s: divide by 3.6 | - | - |

## What's next

Now you have mastered the core content for Movement and Position, you are ready to move on to the next sub-topic in Forces and Motion: Forces, Weight and Newton's Laws. This next section builds directly on the motion calculations you have learned, linking acceleration to the forces acting on an object. You will also apply the motion formulas and graph analysis skills you have practiced here to a wide range of exam questions, including extended response questions and practical scenario questions. Make sure you memorize all three required motion formulas, as they are not provided on the exam formula sheet and will be used repeatedly across the entire Forces and Motion unit. It is also recommended that you practice past paper questions on this topic to familiarize yourself with common exam phrasing and question structures.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/edexcel-igcse-physics-s1-movement-and-position/