Northern IrelandFurther MathsSyllabus dot point

How do you use the constant-acceleration equations and motion graphs to describe motion in a straight line?

Model motion in a straight line with constant acceleration: use the suvat equations and interpret displacement-time and velocity-time graphs.

A CCEA GCSE Further Mathematics answer on kinematics with constant acceleration, covering the suvat equations, vertical motion under gravity, and reading displacement-time and velocity-time graphs in the Mechanics unit.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

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
The suvat equations
A consistent positive direction
Vertical motion under gravity
Motion graphs
Why this matters

What this dot point is asking

Kinematics is the description of motion without worrying about its causes, and CCEA GCSE Further Mathematics begins the Mechanics unit with motion in a straight line at constant acceleration. You must use the four constant-acceleration equations (the suvat equations), apply them to vertical motion under gravity, and read information from displacement-time and velocity-time graphs. Setting out the known quantities clearly is the habit that earns marks reliably.

The suvat equations

The four equations connect the five quantities of constant-acceleration motion, with each equation missing exactly one of them. List the values you know, decide what you want, then choose the equation that contains those and avoids the unknown you do not need.

Choosing the right equation is the skill. If you know three of the five quantities, one equation will give a fourth in a single step, which is why writing the list of suvat values first is so valuable.

A consistent positive direction

Velocity, acceleration and displacement are directed quantities, so you must fix a positive direction at the start and stick with it. Anything in the opposite direction is negative.

Vertical motion under gravity

An object moving vertically under gravity alone has constant acceleration $g$ directed downward, about $9.8$ m/s $^2$ . Treat it as a suvat problem with $a = g$ , choosing signs to match your positive direction.

Vertical motion

A stone is dropped from rest from a cliff and hits the sea after $3$ seconds. Taking $g = 9.8$ m/s $^2$ and downward as positive, find the height of the cliff and the speed on impact.

step 1 List the values

Dropped from rest means $u = 0$ . With downward positive, $a = 9.8$ and $t = 3$ ; we want $s$ and $v$ .

step 2 Find the height

Use $s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(9.8)(3^2) = 4.9 \times 9 = 44.1$ m.

step 3 Find the impact speed

Use $v = u + at = 0 + 9.8 \times 3 = 29.4$ m/s.

step 4 State the answers

The cliff is $44.1$ m high and the stone hits the sea at $29.4$ m/s downward. Both follow directly once the values are listed.

Motion graphs

Graphs give a second route to the same information. On a displacement-time graph the gradient at a point is the velocity, so a steeper line means faster motion and a horizontal line means rest. On a velocity-time graph the gradient is the acceleration and the area between the line and the time axis is the displacement, found as the area of triangles and rectangles. A straight, sloping velocity-time line is exactly the constant-acceleration case, which is why the area gives the same displacement as $s = \tfrac{(u + v)}{2}t$ , the area of a trapezium.

Why this matters

Constant-acceleration kinematics is the foundation of the whole Mechanics unit. The suvat equations reappear in projectile motion, where horizontal and vertical components are each treated as straight-line motion, and they connect to forces through Newton's second law, which supplies the acceleration. The graph methods also link back to calculus, since gradient and area are differentiation and integration in disguise.

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 Unit 2 (style)4 marksA car accelerates uniformly from

8

m/s to

20

m/s in

6

seconds. Find its acceleration and the distance travelled.

Show worked answer →

List the values: $u = 8$ , $v = 20$ , $t = 6$ , with $a$ and $s$ unknown.

Acceleration: $a = \dfrac{v - u}{t} = \dfrac{20 - 8}{6} = 2$ m/s $^2$ .

Distance: use $s = \dfrac{(u + v)}{2}t = \dfrac{(8 + 20)}{2}(6) = 14 \times 6 = 84$ m.

Marks are for choosing suitable equations, the acceleration, and the distance. Listing the suvat values first prevents picking an equation with two unknowns.

CCEA Unit 2 (style)4 marksA ball is thrown vertically upwards at

21

m/s. Taking

g = 9.8

m/s

^2

, find the greatest height reached.

Show worked answer →

At the greatest height the velocity is momentarily zero, so $v = 0$ , $u = 21$ and $a = -9.8$ (gravity acts downward, opposing the motion).

Use $v^2 = u^2 + 2as$ : $0 = 21^2 + 2(-9.8)s$ , so $0 = 441 - 19.6s$ .

Therefore $19.6s = 441$ , giving $s = 22.5$ m.

Marks are for setting $v = 0$ at the top, the negative acceleration, and the height. Taking $a$ as positive is the usual error.

Related dot points

Sources & how we know this

CCEA GCSE Further Mathematics specification (2330) — CCEA (2017)