A runner speeds up from 2\,m/s to 8\,m/s in 3\,s. Calculate the acceleration. [2 marks]

A ball is dropped from rest and falls for 2\,s. Using g = 10\,m/s^2, calculate its velocity just before it lands. [2 marks]

Pearson Edexcel Edexcel 2019-style practice: A train accelerates uniformly from a velocity of 6\,m/s to a velocity of 30\,m/s in a time of 40\,s. Calculate the acceleration of the train, and state the unit.

Use the acceleration equation a = v - ut with v = 30\,m/s, u = 6\,m/s and t = 40\,s (1 mark for selecting the equation). Substitute: a = 30 - 640 = 2440 (1 mark for the change in velocity), giving a = 0.6\,m/s^2 (1 mark) with the unit m/s^2 (1 mark). Markers reward using the change in velocity (not the final velocity), correct substitution and the squared unit. A common error is to divide 30 by 40 and forget to subtract the initial velocity.

Pearson Edexcel Edexcel 2021-style practice: A car travelling at 8\,m/s accelerates at 2\,m/s^2 over a distance of 36\,m. Calculate the final velocity of the car. Use the equation v^2 - u^2 = 2 × a × x.

Substitute u = 8\,m/s, a = 2\,m/s^2 and x = 36\,m into v^2 - u^2 = 2ax: v^2 = u^2 + 2ax = 8^2 + 2 × 2 × 36 = 64 + 144 = 208 (2 marks for correct substitution and evaluation). Take the square root: v = sqrt(208) = 14.4\,m/s (1 mark). Markers reward rearranging to make v^2 the subject, adding u^2 rather than ignoring it, and remembering the final square root. Leaving the answer as 208 (forgetting the root) is the most common slip.

EnglandPhysicsSyllabus dot point

How do you calculate acceleration, and how do the equations of motion link velocity, acceleration and distance?

Acceleration and the equations of motion: the acceleration equation, the uniform acceleration (suvat) equation linking velocity, acceleration and distance, and typical accelerations such as g.

A focused answer to Edexcel GCSE Physics 2.8 and 2.9, covering the acceleration equation, the uniform acceleration equation linking final velocity, initial velocity, acceleration and distance, the meaning of negative acceleration, and the acceleration of free fall, with worked calculations in the Edexcel style.

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

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

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What this dot point is asking
The acceleration equation
Typical accelerations and free fall
The uniform acceleration equation
Try this

What this dot point is asking

Edexcel statements 2.8 and 2.9 want you to recall and use the acceleration equation, and to use the uniform acceleration equation that links final velocity, initial velocity, acceleration and distance. You also need to recall that the acceleration of free fall near the Earth is roughly $10\,\text{m/s}^2$ and that a negative acceleration means slowing down.

The acceleration equation

The quantity $v - u$ is the change in velocity, sometimes written $\Delta v$ . Because velocity is a vector, a change in direction is also a change in velocity, even if the speed stays the same. If the final velocity is smaller than the initial velocity the answer is negative, which Edexcel calls a deceleration: the object is slowing down. Always substitute in SI units, so convert any speed given in $\text{km/h}$ to $\text{m/s}$ first.

Typical accelerations and free fall

Edexcel expects you to have a feel for the size of everyday accelerations: a family car pulling away reaches perhaps $2$ to $3\,\text{m/s}^2$ , while $g$ is about $10\,\text{m/s}^2$ . Knowing the rough size lets you spot an answer that is wildly wrong, for example an acceleration of $400\,\text{m/s}^2$ for a car, which usually signals a missing unit conversion.

The uniform acceleration equation

This equation (the only one of the "suvat" family Edexcel uses) is the tool to reach for when a question gives a distance but no time. It links four quantities, so given any three you can find the fourth. Rearrange carefully: to find the final velocity, make $v^2$ the subject as $v^2 = u^2 + 2ax$ and take the square root at the end. Watch the sign of $a$ : for a decelerating object (such as a braking car) $a$ is negative, so $2ax$ is subtracted and $v$ comes out smaller than $u$ .

Worked example: stopping distance from a deceleration

A cyclist moving at $u = 12\,\text{m/s}$ brakes uniformly and stops. The deceleration is $3\,\text{m/s}^2$ . Find the stopping distance.

Step 1: Identify the known quantities and the missing one

We know $u = 12\,\text{m/s}$ (initial velocity), $v = 0\,\text{m/s}$ (final velocity, the cyclist stops), and $a = -3\,\text{m/s}^2$ (deceleration, so the acceleration is negative). The question asks for the distance $x$ , and no time is given. This tells us to reach for the equation that links velocity, acceleration and distance, without needing time.

Step 2: Choose the equation and rearrange for $x$

The uniform acceleration equation $v^2 - u^2 = 2ax$ contains all four quantities. Rearranging for $x$ :

x = \frac{v^2 - u^2}{2a}

Step 3: Substitute and calculate

x = \frac{0^2 - 12^2}{2 \times (-3)} = \frac{-144}{-6} = 24\,\text{m}

The two negatives cancel, giving a positive distance, which makes physical sense.

Final answer: The stopping distance is $24\,\text{m}$ .

Try this

Q1. A runner speeds up from $2\,\text{m/s}$ to $8\,\text{m/s}$ in $3\,\text{s}$ . Calculate the acceleration. [2 marks]

Cue. $a = \dfrac{v - u}{t} = \dfrac{8 - 2}{3} = 2\,\text{m/s}^2$ .

Q2. A ball is dropped from rest and falls for $2\,\text{s}$ . Using $g = 10\,\text{m/s}^2$ , calculate its velocity just before it lands. [2 marks]

Cue. Rearrange $a = \dfrac{v - u}{t}$ to $v = u + at = 0 + 10 \times 2 = 20\,\text{m/s}$ .

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 train accelerates uniformly from a velocity of

6\,\text{m/s}

to a velocity of

30\,\text{m/s}

in a time of

40\,\text{s}

. Calculate the acceleration of the train, and state the unit.

Show worked answer →

Use the acceleration equation $a = \dfrac{v - u}{t}$ with $v = 30\,\text{m/s}$ , $u = 6\,\text{m/s}$ and $t = 40\,\text{s}$ (1 mark for selecting the equation). Substitute: $a = \dfrac{30 - 6}{40} = \dfrac{24}{40}$ (1 mark for the change in velocity), giving $a = 0.6\,\text{m/s}^2$ (1 mark) with the unit $\text{m/s}^2$ (1 mark). Markers reward using the change in velocity (not the final velocity), correct substitution and the squared unit. A common error is to divide $30$ by $40$ and forget to subtract the initial velocity.

Edexcel 20213 marksA car travelling at

8\,\text{m/s}

accelerates at

2\,\text{m/s}^2

over a distance of

36\,\text{m}

. Calculate the final velocity of the car. Use the equation

v^2 - u^2 = 2 \times a \times x

Show worked answer →

Substitute $u = 8\,\text{m/s}$ , $a = 2\,\text{m/s}^2$ and $x = 36\,\text{m}$ into $v^2 - u^2 = 2ax$ : $v^2 = u^2 + 2ax = 8^2 + 2 \times 2 \times 36 = 64 + 144 = 208$ (2 marks for correct substitution and evaluation). Take the square root: $v = \sqrt{208} = 14.4\,\text{m/s}$ (1 mark). Markers reward rearranging to make $v^2$ the subject, adding $u^2$ rather than ignoring it, and remembering the final square root. Leaving the answer as $208$ (forgetting the root) is the most common slip.

Related dot points

Sources & how we know this

Pearson Edexcel GCSE (9-1) Physics (1PH0) specification — Pearson (2016)
Edexcel GCSE Physics and Combined Science equation list (1PH0/1SC0) — Pearson (2025)