What is sign errors under gravity?

Decide a positive direction once and stick to it; with up positive, the acceleration is -9.8\,m s^-2 and a downward final velocity is negative.

A car travels at constant 15\,m s^-1 for 12\,s. Find the distance. [1 mark]

A ball is dropped from rest. Find its speed after 2\,s, taking g = 9.8\,m s^-2. [2 marks]

CCEA CCEA 2019-style practice: A stone is projected vertically upwards from ground level with a speed of 21\,m s^-1. Taking g = 9.8\,m s^-2, find the greatest height reached and the total time before it returns to the ground.

Take upwards as positive, so a = -9.8\,m s^-2, u = 21\,m s^-1. At the greatest height v = 0. Using v^2 = u^2 + 2as: 0 = 21^2 + 2(-9.8)s s = (441)/(19.6) = 22.5\,m. For the total time, the stone returns to s = 0. Using s = ut + (1)/(2)at^2: 0 = 21t - 4.9t^2 = t(21 - 4.9t), so t = 0 (start) or t = (21)/(4.9) = 4.29\,s. The greatest height is 22.5\,m and the total time is 4.29\,s. Markers reward the sign convention, the use of v = 0 at the top, the height, and the total time from the displacement equation.

CCEA CCEA 2021-style practice: A car accelerates uniformly from 8\,m s^-1 to 20\,m s^-1 over a distance of 84\,m. Find the acceleration and the time taken.

Use v^2 = u^2 + 2as with u = 8, v = 20, s = 84: 20^2 = 8^2 + 2a(84) 400 = 64 + 168a a = (336)/(168) = 2\,m s^-2. For the time use v = u + at: 20 = 8 + 2t t = 6\,s. Markers reward selecting the equation without time first, the acceleration, then the time from v = u + at.

Northern IrelandMathsSyllabus dot point

How do the equations of motion and motion graphs describe a body moving in a straight line?

Displacement, velocity and acceleration for motion in a straight line, the equations of motion for constant acceleration, motion under gravity, and interpreting displacement-time and velocity-time graphs.

A CCEA A-Level Mathematics answer on displacement, velocity and acceleration in a straight line, the suvat equations of motion for constant acceleration, motion under gravity, and reading and using displacement-time and velocity-time graphs.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

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

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What this dot point is asking

CCEA wants you to define displacement, velocity and acceleration for motion in a straight line, apply the equations of motion (the suvat equations) for constant acceleration, solve problems of motion under gravity, and read and use displacement-time and velocity-time graphs. Kinematics is the opening of the mechanics half of AS 2 and feeds directly into the work on forces.

The answer

Displacement, velocity and acceleration

The equations of motion

To choose an equation, list the quantities you know and want, then pick the one missing the variable you neither know nor need.

Motion under gravity

A body moving freely under gravity has constant acceleration $g \approx 9.8\,\text{m s}^{-2}$ downwards (air resistance ignored). Choose a positive direction and keep signs consistent: if up is positive, then $a = -g$ . At the highest point of a vertical throw the velocity is momentarily zero, which is the key fact for finding the maximum height.

Motion graphs

Worked example: a two-stage journey

Find a total distance from a velocity-time graph

A train accelerates uniformly from rest to $30\,\text{m s}^{-1}$ in $20\,\text{s}$ , travels at $30\,\text{m s}^{-1}$ for $60\,\text{s}$ , then decelerates uniformly to rest in $15\,\text{s}$ . Find the total distance.

step Acceleration phase

The area is a triangle: $\frac{1}{2}(20)(30) = 300\,\text{m}$ .

step Constant-speed phase

The area is a rectangle: $30 \times 60 = 1800\,\text{m}$ .

step Deceleration phase and total

The area is a triangle: $\frac{1}{2}(15)(30) = 225\,\text{m}$ .

Adding: $300 + 1800 + 225 = 2325\,\text{m}$ . The area under the velocity-time graph gives the displacement for each phase.

Examples in context

Example 1. Stopping distance. A driver braking from $30\,\text{m s}^{-1}$ at $5\,\text{m s}^{-2}$ stops in $v^2 = u^2 + 2as$ with $v = 0$ : $0 = 900 - 10s$ , so $s = 90\,\text{m}$ . The suvat equations give the road-safety stopping distance directly.

Example 2. A lift's motion. A lift speeding up, cruising and slowing down traces a trapezium on a velocity-time graph. The total distance is the trapezium's area, and the acceleration phases are the sloping sides, exactly the structure of the worked example above.

Try this

Q1. A car travels at constant $15\,\text{m s}^{-1}$ for $12\,\text{s}$ . Find the distance. [1 mark]

Cue. $s = 15 \times 12 = 180\,\text{m}$ .

Q2. A ball is dropped from rest. Find its speed after $2\,\text{s}$ , taking $g = 9.8\,\text{m s}^{-2}$ . [2 marks]

Cue. $v = u + at = 0 + 9.8(2) = 19.6\,\text{m s}^{-1}$ .

Q3. On a velocity-time graph, what does the area under the line represent? [1 mark]

Cue. The displacement.

Exam-style practice questions

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

CCEA 20196 marksA stone is projected vertically upwards from ground level with a speed of

21\,\text{m s}^{-1}

. Taking

g = 9.8\,\text{m s}^{-2}

, find the greatest height reached and the total time before it returns to the ground.

Show worked answer →

Take upwards as positive, so $a = -9.8\,\text{m s}^{-2}$ , $u = 21\,\text{m s}^{-1}$ .

At the greatest height $v = 0$ . Using $v^2 = u^2 + 2as$ :

$0 = 21^2 + 2(-9.8)s \Rightarrow s = \frac{441}{19.6} = 22.5\,\text{m}.$

For the total time, the stone returns to $s = 0$ . Using $s = ut + \frac{1}{2}at^2$ :

$0 = 21t - 4.9t^2 = t(21 - 4.9t),$

so $t = 0$ (start) or $t = \frac{21}{4.9} = 4.29\,\text{s}$ .

The greatest height is $22.5\,\text{m}$ and the total time is $4.29\,\text{s}$ .

Markers reward the sign convention, the use of $v = 0$ at the top, the height, and the total time from the displacement equation.

CCEA 20215 marksA car accelerates uniformly from

8\,\text{m s}^{-1}

20\,\text{m s}^{-1}

over a distance of

84\,\text{m}

. Find the acceleration and the time taken.

Show worked answer →

Use $v^2 = u^2 + 2as$ with $u = 8$ , $v = 20$ , $s = 84$ :

$20^2 = 8^2 + 2a(84) \Rightarrow 400 = 64 + 168a \Rightarrow a = \frac{336}{168} = 2\,\text{m s}^{-2}.$

For the time use $v = u + at$ :

$20 = 8 + 2t \Rightarrow t = 6\,\text{s}.$

Markers reward selecting the equation without time first, the acceleration, then the time from $v = u + at$ .

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)