Pearson Edexcel Edexcel 2019-style practice: A car accelerates uniformly from rest to a velocity of 24\,m/s in 8.0\,s. Calculate its acceleration, and then the distance it travels in this time.

A 4-mark calculation using two equations. Acceleration a = vt = 24 - 08.0 = 3.0\,m/s^2 (2 marks). For the distance, the average velocity is 0 + 242 = 12\,m/s, so distance = average velocity multiplied by time = 12 × 8.0 = 96\,m (2 marks). Alternatively, use v^2 - u^2 = 2as rearranged for s. Markers reward the acceleration with units, and the distance using average velocity times time (or the uniform acceleration equation).

EnglandCombined ScienceSyllabus dot point

How do we describe and calculate motion using speed, velocity and acceleration?

Distance, displacement, speed and velocity, the equations for speed and acceleration, the uniform acceleration equation, and interpreting distance-time and velocity-time graphs.

A focused answer to Edexcel GCSE Combined Science Topic 2 (CP2), covering distance, displacement, speed and velocity, the equations for speed and acceleration, the uniform acceleration equation, and how to interpret distance-time and velocity-time graphs.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Distance, displacement, speed and velocity
The equations
Worked calculation
Motion graphs
Try this

What this dot point is asking

Edexcel wants you to distinguish distance and displacement and speed and velocity, use the equations for speed and acceleration, apply the uniform acceleration equation, and interpret distance-time and velocity-time graphs.

Distance, displacement, speed and velocity

Typical speeds to know: walking about $1.5\,m/s$ , running about $3\,m/s$ , cycling about $6\,m/s$ , and the speed of sound in air about $330\,m/s$ . Knowing these typical values lets you check whether an answer is sensible, and they sometimes appear directly in exam questions, for example estimating how long a journey takes at walking pace.

The equations

Here $u$ is the starting velocity, $v$ the final velocity, $a$ the acceleration in $m/s^2$ , and $t$ the time in seconds. A positive acceleration means speeding up; a negative acceleration (a deceleration) means slowing down. If an object starts from rest, its starting velocity $u$ is zero, which simplifies the equations. The uniform (constant) acceleration equation links velocity, acceleration and distance:

v^2 - u^2 = 2as

Worked calculation

Motion graphs

A distance-time graph shows how distance changes with time:

A horizontal line means the object is stationary.
A straight sloped line means constant speed; the steeper the slope, the faster.
The gradient equals the speed.

A velocity-time graph shows how velocity changes with time:

A horizontal line means constant velocity.
A sloped line means acceleration; the gradient equals the acceleration.
The area under the line equals the distance travelled.

To find the gradient of a straight section, divide the change in the quantity on the vertical axis by the change in time. To find the area under a velocity-time graph, split the shape into rectangles and triangles, work out each area, and add them up. A common exam task is a journey graph with several stages (speeding up, constant speed, slowing down): you read the acceleration from each gradient and the total distance from the total area. A negative gradient on a velocity-time graph means the object is decelerating (slowing down).

Acceleration due to gravity near the Earth's surface is about $10\,m/s^2$ , so a freely falling object speeds up by about $10\,m/s$ every second (ignoring air resistance). This is why heavier and lighter objects fall together in the absence of air resistance: the acceleration does not depend on the mass.

Try this

Q1. State the equation linking speed, distance and time. [1 mark]

Cue. Speed = distance / time.

Q2. A runner travels $100\,m$ in $20\,s$ . Calculate the average speed. [2 marks]

Cue. Speed $= 100 / 20 = 5\,m/s$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksA car accelerates uniformly from rest to a velocity of

24\,\text{m/s}

8.0\,\text{s}

. Calculate its acceleration, and then the distance it travels in this time.

Show worked answer →

A 4-mark calculation using two equations.

Acceleration $a = \dfrac{\Delta v}{t} = \dfrac{24 - 0}{8.0} = 3.0\,\text{m/s}^2$ (2 marks). For the distance, the average velocity is $\dfrac{0 + 24}{2} = 12\,\text{m/s}$ , so distance = average velocity multiplied by time $= 12 \times 8.0 = 96\,\text{m}$ (2 marks). Alternatively, use $v^2 - u^2 = 2as$ rearranged for $s$ .

Markers reward the acceleration with units, and the distance using average velocity times time (or the uniform acceleration equation).

Edexcel 20214 marksDescribe what the gradient and the area under a velocity-time graph represent, and state how you would find the distance travelled from such a graph.

Show worked answer →

A 4-mark question on interpreting velocity-time graphs.

The gradient (slope) of a velocity-time graph represents the acceleration: a steeper line means a greater acceleration, and a horizontal line means constant velocity (zero acceleration) (2 marks). The area under the line represents the distance travelled (1 mark). To find the distance, calculate the area between the line and the time axis, for example by splitting it into rectangles and triangles and adding them up (1 mark).

Markers reward gradient as acceleration, area as distance, and a valid method for finding the area.

Related dot points

Sources & how we know this

Edexcel GCSE (9-1) Combined Science (1SC0) specification — Pearson (2016)