What are displacement-time graphs?

Confusing the two graph types. On a velocity-time graph the area is displacement and the gradient is acceleration; on a displacement-time graph the gradient is velocity. Reading the wrong quantity is a frequent error.

A particle starts at 3\,m s^-1 and accelerates at 2\,m s^-2 for 4\,s. Find its final velocity. [2 marks]

A stone is dropped from rest. Taking g = 9.8\,m s^-2, find its speed after 2\,s. [2 marks]

CCEA CCEA AS 2021-style practice: A car accelerates uniformly from rest to 24\,m s^-1 in 12\,s, then travels at constant speed for 20\,s. Find the acceleration during the first stage and the total distance travelled.

Stage 1, from rest (u = 0) to v = 24\,m s^-1 in t = 12\,s. Using v = u + at: 24 = 0 + a(12) a = 2\,m s^-2. Distance in stage 1, using s = (1)/(2)(u + v)t: s_1 = (1)/(2)(0 + 24)(12) = 144\,m. Stage 2, constant speed 24\,m s^-1 for 20\,s: s_2 = 24 × 20 = 480\,m. Total distance = 144 + 480 = 624\,m. Markers reward the acceleration, the correct suvat choice for each stage, and the total distance.

CCEA CCEA AS 2019-style practice: A ball is thrown vertically upwards with speed 14\,m s^-1 from ground level. Taking g = 9.8\,m s^-2, find the greatest height reached and the time taken to return to the ground.

Take upwards as positive, so a = -9.8\,m s^-2 and u = 14\,m s^-1. At the greatest height v = 0. Using v^2 = u^2 + 2as: 0 = 14^2 + 2(-9.8)s s = 19619.6 = 10\,m. For the time to return to the ground, the displacement is s = 0. Using s = ut + (1)/(2)at^2: 0 = 14t - 4.9t^2 = t(14 - 4.9t), so t = 0 (launch) or t = 144.9 ≈ 2.9\,s. Markers reward the sign convention, the greatest height of 10\,m, and the total time of about 2.9\,s.

Northern IrelandFurther MathsSyllabus dot point

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

Kinematics of motion in a straight line with constant acceleration, the suvat equations, vertical motion under gravity, and interpreting displacement-time and velocity-time graphs.

A CCEA AS Further Maths Mechanics answer on motion in a straight line with constant acceleration, the suvat equations, vertical motion under gravity, and reading displacement-time and velocity-time graphs for gradient and area.

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
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Examples in context
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What this dot point is asking

CCEA wants you to model motion in a straight line with constant acceleration, select and apply the right suvat equation, handle vertical motion under gravity with a clear sign convention, and read displacement-time and velocity-time graphs, where gradient and area carry physical meaning.

The answer

The suvat equations

Vertical motion under gravity

Velocity-time graphs

Displacement-time graphs

Reading a velocity-time graph

A cyclist accelerates uniformly from $4\,\text{m s}^{-1}$ to $10\,\text{m s}^{-1}$ over $6\,\text{s}$ , then decelerates uniformly to rest in a further $5\,\text{s}$ . Find the total distance using the graph.

step Distance in the acceleration phase

This is a trapezium of parallel sides $4$ and $10$ and width $6$ :

s_1 = \tfrac{1}{2}(4 + 10)(6) = \tfrac{1}{2}(14)(6) = 42\,\text{m}.

step Distance in the deceleration phase

A triangle of height $10$ and base $5$ :

s_2 = \tfrac{1}{2}(10)(5) = 25\,\text{m}.

step Add the areas

s = 42 + 25 = 67\,\text{m}.

The total displacement is the total area under the velocity-time graph.

Examples in context

Example 1. Stopping distances in the Highway Code. The braking distance of a car comes straight from $v^2 = u^2 + 2as$ : with $v = 0$ , the distance grows with the square of the initial speed, which is why doubling speed roughly quadruples the braking distance.

Example 2. Reading a train's data recorder. A train's velocity-time trace lets engineers find the distance between stations as the area under the curve and the braking rate as the gradient of the final section. The graph encodes both at once.

Try this

Q1. A particle starts at $3\,\text{m s}^{-1}$ and accelerates at $2\,\text{m s}^{-2}$ for $4\,\text{s}$ . Find its final velocity. [2 marks]

Cue. $v = u + at = 3 + 2(4) = 11\,\text{m s}^{-1}$ .

Q2. What does the area under a velocity-time graph represent? [1 mark]

Cue. The displacement.

Q3. A stone is dropped from rest. Taking $g = 9.8\,\text{m s}^{-2}$ , find its speed after $2\,\text{s}$ . [2 marks]

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

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 AS 20216 marksA car accelerates uniformly from rest to

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

12\,\text{s}

, then travels at constant speed for

20\,\text{s}

. Find the acceleration during the first stage and the total distance travelled.

Show worked answer →

Stage 1, from rest ( $u = 0$ ) to $v = 24\,\text{m s}^{-1}$ in $t = 12\,\text{s}$ . Using $v = u + at$ :

$24 = 0 + a(12) \Rightarrow a = 2\,\text{m s}^{-2}.$

Distance in stage 1, using $s = \frac{1}{2}(u + v)t$ :

$s_1 = \frac{1}{2}(0 + 24)(12) = 144\,\text{m}.$

Stage 2, constant speed $24\,\text{m s}^{-1}$ for $20\,\text{s}$ :

$s_2 = 24 \times 20 = 480\,\text{m}.$

Total distance $= 144 + 480 = 624\,\text{m}.$

Markers reward the acceleration, the correct suvat choice for each stage, and the total distance.

CCEA AS 20195 marksA ball is thrown vertically upwards with speed

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

from ground level. Taking

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

, find the greatest height reached and the time taken to return to the ground.

Show worked answer →

Take upwards as positive, so $a = -9.8\,\text{m s}^{-2}$ and $u = 14\,\text{m s}^{-1}$ .

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

$0 = 14^2 + 2(-9.8)s \Rightarrow s = \dfrac{196}{19.6} = 10\,\text{m}.$

For the time to return to the ground, the displacement is $s = 0$ . Using $s = ut + \frac{1}{2}at^2$ :

$0 = 14t - 4.9t^2 = t(14 - 4.9t),$ so $t = 0$ (launch) or $t = \dfrac{14}{4.9} \approx 2.9\,\text{s}.$

Markers reward the sign convention, the greatest height of $10\,\text{m}$ , and the total time of about $2.9\,\text{s}$ .

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)