Acceleration and equations of motion

What you'll learn

This guide covers acceleration and the equations of motion as required by the AQA GCSE Physics specification. You will learn how to calculate acceleration from velocity changes, apply the key equations of motion to solve problems involving uniform acceleration, and interpret velocity-time graphs. These concepts are essential for both Foundation and Higher tier papers and form the basis of many multi-step calculation questions.

Key terms and definitions

Acceleration — the rate of change of velocity, measured in metres per second squared (m/s²). Acceleration can be positive (speeding up), negative/deceleration (slowing down), or involve a change in direction.

Velocity — the speed of an object in a specific direction, measured in metres per second (m/s). Unlike speed, velocity is a vector quantity.

Uniform acceleration — constant acceleration where the velocity changes by the same amount every second. Most GCSE questions assume uniform acceleration unless stated otherwise.

Displacement — the distance travelled in a straight line from the starting position to the finishing position, measured in metres (m). This is a vector quantity with both magnitude and direction.

Initial velocity — the velocity at which an object starts moving, represented by the symbol u in equations of motion (m/s).

Final velocity — the velocity at which an object finishes moving, represented by the symbol v in equations of motion (m/s).

Deceleration — negative acceleration where an object slows down. The velocity decreases over time, so acceleration has a negative value.

Core concepts

Calculating acceleration

Acceleration is calculated using the equation:

a = (v - u) / t

Where:

• a = acceleration (m/s²)
• v = final velocity (m/s)
• u = initial velocity (m/s)
• t = time taken (s)

This equation appears on the AQA Physics equation sheet, so you don't need to memorise it, but you must understand how to rearrange and apply it.

Key points about acceleration:

• If the final velocity is greater than initial velocity, acceleration is positive
• If the final velocity is less than initial velocity, acceleration is negative (deceleration)
• An object moving at constant velocity has zero acceleration
• Acceleration is a vector quantity — it has both magnitude and direction
• The unit m/s² means "metres per second per second" or "metres per second, per second"

The equations of motion

There are four main equations of motion used in GCSE Physics. These apply only to objects moving with uniform acceleration in a straight line.

Equation 1: v = u + at

This relates final velocity to initial velocity, acceleration and time.

Equation 2: s = ((u + v) / 2) × t

This calculates displacement using average velocity multiplied by time. The term (u + v) / 2 represents the average velocity.

Equation 3: s = ut + ½at²

This calculates displacement when you know initial velocity, time and acceleration but not final velocity.

Equation 4: v² = u² + 2as

This relates velocities, acceleration and displacement without needing to know the time taken.

All distances in these equations refer to displacement (s), not total distance travelled. The standard symbols are:

• s = displacement (m)
• u = initial velocity (m/s)
• v = final velocity (m/s)
• a = acceleration (m/s²)
• t = time (s)

Choosing the right equation

You must select the appropriate equation based on which variables are given and which you need to find:

Strategy for solving problems:

1. Write down what you know (including any zero values)
2. Identify what you need to find
3. Select the equation containing these four quantities
4. Substitute values carefully, including units
5. Rearrange if necessary
6. Calculate and include the correct unit in your answer

Velocity-time graphs

Velocity-time graphs are powerful tools for analysing motion with uniform acceleration. You must be able to interpret and draw these graphs.

Gradient of a velocity-time graph:

• The gradient (slope) represents acceleration
• Steeper gradient = greater acceleration
• Horizontal line (zero gradient) = constant velocity (zero acceleration)
• Negative gradient = deceleration
• Calculate gradient using: acceleration = change in velocity / change in time

Area under a velocity-time graph:

• The area between the line and the time axis represents displacement
• For a triangle: area = ½ × base × height
• For a rectangle: area = length × width
• For complex shapes, split into triangles and rectangles
• Area below the time axis (negative velocity) represents displacement in the opposite direction

Common graph features:

• Straight diagonal line = uniform acceleration
• Curved line = non-uniform acceleration (not typically tested at GCSE)
• The y-intercept shows the initial velocity
• Where the line crosses the time axis, velocity = 0

Acceleration due to gravity

Near Earth's surface, all objects in free fall experience the same acceleration due to gravity (assuming air resistance is negligible).

g = 9.8 m/s² (or approximately 10 m/s² for easier calculations)

When using equations of motion for objects falling or thrown vertically:

• Use g as the value for acceleration (a = g = 9.8 m/s²)
• For objects falling downwards, g is positive
• For objects thrown upwards, g is negative (acts against the motion)
• At the highest point of vertical motion, velocity = 0 (v = 0)
• Air resistance is usually ignored in GCSE calculations unless specifically mentioned

Typical motion scenarios

Accelerating from rest:

• Initial velocity u = 0
• Equations simplify: v = at, s = ½at², v² = 2as

Coming to rest (stopping):

• Final velocity v = 0
• Useful for braking distance calculations

Constant velocity:

• Acceleration a = 0
• Initial velocity equals final velocity (u = v)
• Use distance = speed × time instead

Vertical motion upwards:

• Acceleration is negative (a = -9.8 m/s²)
• At maximum height, v = 0
• Time to rise equals time to fall (in the absence of air resistance)

Worked examples

Example 1: Basic acceleration calculation

Question: A car accelerates from rest to 28 m/s in 8.0 seconds. Calculate the acceleration of the car. (3 marks)

Solution:

• Write the known values: u = 0 m/s (from rest), v = 28 m/s, t = 8.0 s (1 mark)
• Select equation: a = (v - u) / t (1 mark for correct substitution)
• Calculate: a = (28 - 0) / 8.0 = 3.5 m/s² (1 mark)

Mark scheme notes: You must show the equation (or rearranged version), substitution with values, and the final answer with correct unit. Writing u = 0 explicitly demonstrates understanding.

Example 2: Using equations of motion

Question: A cyclist travels with uniform acceleration. She accelerates from 4.0 m/s to 10.0 m/s over a distance of 35 m. Calculate: (a) The acceleration of the cyclist (3 marks) (b) The time taken (3 marks)

Solution:

(a) Finding acceleration:

• Known: u = 4.0 m/s, v = 10.0 m/s, s = 35 m
• Unknown: a (we don't know t, so cannot use a = (v-u)/t)
• Use: v² = u² + 2as (1 mark)
• Rearrange: a = (v² - u²) / 2s
• Substitute: a = (10.0² - 4.0²) / (2 × 35) = (100 - 16) / 70 = 84 / 70 (1 mark)
• a = 1.2 m/s² (1 mark)

(b) Finding time:

• Use: v = u + at (1 mark)
• Rearrange: t = (v - u) / a
• Substitute: t = (10.0 - 4.0) / 1.2 = 6.0 / 1.2 (1 mark)
• t = 5.0 s (1 mark)

Alternative for (b): Could use s = ((u + v) / 2) × t and rearrange.

Example 3: Velocity-time graph interpretation

Question: The graph shows the motion of a train between two stations.

[Graph description: Straight line from (0s, 0m/s) to (20s, 30m/s), horizontal line from (20s, 30m/s) to (50s, 30m/s), straight line from (50s, 30m/s) to (70s, 0m/s)]

(a) Calculate the acceleration during the first 20 seconds. (2 marks) (b) Calculate the total distance travelled by the train. (4 marks)

Solution:

(a) Acceleration = gradient = change in velocity / change in time (1 mark)

• a = (30 - 0) / (20 - 0) = 30 / 20 = 1.5 m/s² (1 mark)

(b) Distance = area under graph

• Section 1 (triangle): ½ × 20 × 30 = 300 m (1 mark)
• Section 2 (rectangle): 30 × (50 - 20) = 30 × 30 = 900 m (1 mark)
• Section 3 (triangle): ½ × 20 × 30 = 300 m (1 mark)
• Total distance = 300 + 900 + 300 = 1500 m (or 1.5 km) (1 mark)

Common mistakes and how to avoid them

• Confusing velocity and speed: Remember velocity has direction. In equations of motion, always use velocity values with consideration for direction. If an object changes direction, velocities in opposite directions have opposite signs.

• Using the wrong equation: List all known variables (including u = 0 if starting from rest, or v = 0 if stopping). Count that you have three known quantities before selecting an equation. The equation must contain the three known values plus the one unknown.

• Unit errors: All values must be in standard SI units before calculation: metres (not km or cm), seconds (not minutes or hours), m/s (not km/h). Convert before substituting. Always include the unit with your final answer.

• Sign errors with deceleration and vertical motion: When an object slows down, acceleration is negative. For objects thrown upwards, use a = -9.8 m/s². Check whether your answer makes physical sense.

• Forgetting to show working: Exam mark schemes award method marks. Always write the equation, show substitution of values, then calculate. Even if your final answer is wrong, you can gain partial credit.

• Velocity-time graph errors: Remember gradient = acceleration (not speed), and area = displacement (not velocity). Use the correct formulas for triangles and rectangles. Don't forget units (m/s² for gradient, m for area).

Exam technique for "Acceleration and equations of motion"

• Command word "Calculate": You must show your working to gain full marks. Write the equation, substitute values, and give your answer with the correct unit. Typical allocation is 1 mark for equation selection, 1 mark for correct substitution/working, 1 mark for answer.

• Standard form and significant figures: Give answers to the same number of significant figures as the data provided (usually 2 or 3). Use standard form for very large or small numbers. Check whether the question specifies the required format.

• Graph questions carry multiple marks: For sketches, label axes with quantities and units. Plot points accurately using a ruler. When calculating gradient or area, show the method clearly with values from the graph marked on.

• Multi-step problems: Break the motion into separate stages. Solve for each stage systematically. Remember the final velocity of one stage becomes the initial velocity of the next stage.

Quick revision summary

Acceleration is the rate of change of velocity (a = (v-u)/t). The four equations of motion link displacement, initial velocity, final velocity, acceleration and time for uniform acceleration. Select equations based on known and unknown variables. On velocity-time graphs, gradient represents acceleration and area represents displacement. Free fall acceleration is 9.8 m/s². Always show full working in calculations, use correct units, and check answers make physical sense. Practice selecting and rearranging equations until confident.