What is sign errors with direction?

Choose a positive direction and keep displacement, velocity and acceleration consistent with it.

EnglandMathsSyllabus dot point

How do you describe and calculate motion using displacement, velocity, acceleration and time?

Displacement, velocity and acceleration, motion graphs, the constant acceleration formulae, and using calculus to relate displacement, velocity and acceleration that vary with time.

A focused answer to the Edexcel A-Level Mathematics kinematics content, covering displacement, velocity and acceleration, motion graphs, the constant acceleration formulae, and using calculus to relate displacement, velocity and acceleration that vary with time.

Generated by Claude Opus 4.810 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 answer
Examples in context
Try this

What this dot point is asking

Edexcel wants you to work with displacement, velocity and acceleration, interpret and draw displacement-time and velocity-time graphs, use the constant acceleration (suvat) formulae, and use differentiation and integration to relate displacement, velocity and acceleration when acceleration is not constant.

The answer

Displacement, velocity and acceleration

Motion graphs

On a displacement-time graph the gradient gives velocity. On a velocity-time graph the gradient gives acceleration and the area under the graph gives displacement. These let you read motion straight from a sketch. A horizontal line on a displacement-time graph means the body is stationary, while a horizontal line on a velocity-time graph means constant velocity. A curve on a velocity-time graph signals non-constant acceleration, which is the clue to switch from suvat to calculus. When a velocity-time graph dips below the axis the body is moving in the negative direction, so the area there counts as negative displacement; to find the total distance travelled you add the magnitudes of the areas above and below the axis separately.

Constant acceleration

Variable acceleration

When acceleration varies, use calculus. If displacement is $s(t)$ , then velocity is $v = \dfrac{ds}{dt}$ and acceleration is $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$ . Reversing, $s = \int v\,dt$ and $v = \int a\,dt$ , with constants found from initial conditions. A turning point of the displacement happens where $v = 0$ , and a maximum or minimum velocity happens where $a = 0$ , so calculus also locates the moments when the body is instantaneously at rest or moving fastest.

Examples in context

A particle moves with velocity

v = 12 - 2t \text{ m s}^{-1}

. Find the distance travelled in the first

8

step 1 Find when the velocity is zero

$12 - 2t = 0$ gives $t = 6$ s, so the particle reverses direction at $t = 6$ .

step 2 Distance forward from $t = 0$ to $t = 6$

$\int_0^6 (12 - 2t)\,dt = [12t - t^2]_0^6 = 72 - 36 = 36$ m.

step 3 Distance back from $t = 6$ to $t = 8$

$\int_6^8 (12 - 2t)\,dt = [12t - t^2]_6^8 = (96 - 64) - (72 - 36) = 32 - 36 = -4$ m, a distance of $4$ m.

step 4 Total distance

$36 + 4 = 40$ m, while the net displacement is only $36 - 4 = 32$ m.

Try this

Q1. A ball is thrown up at $20$ m per second. Taking $g = 9.8$ , find the time to reach its highest point. [3 marks]

Cue. At the top $v = 0$ , so $0 = 20 - 9.8t$ gives $t \approx 2.04$ s.

Q2. A particle has velocity $v = 3t^2$ m per second. Find its acceleration at $t = 2$ . [2 marks]

Cue. $a = \dfrac{dv}{dt} = 6t = 12$ m per second squared.

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 20186 marksA particle moves in a straight line. It passes a point with velocity

4

m s

^{-1}

and decelerates uniformly, coming to rest after travelling

20

m. Find the deceleration and the time taken.

Show worked answer →

Use $v^2 = u^2 + 2as$ with $u = 4$ , $v = 0$ , $s = 20$ (M1): $0 = 16 + 2a(20)$ , so $40a = -16$ and $a = -0.4$ m s $^{-2}$ (A1).

The deceleration is $0.4$ m s $^{-2}$ (A1).

Use $v = u + at$ (M1): $0 = 4 + (-0.4)t$ , so $t = \dfrac{4}{0.4} = 10$ s (A1).

Alternatively $s = \dfrac{1}{2}(u + v)t$ gives $20 = \dfrac{1}{2}(4)t$ , so $t = 10$ s (A1).

Markers reward the correct suvat choice, the deceleration, and the time.

Edexcel 20216 marksA particle moves so that its velocity is

v = 3t^2 - 12t + 9

m s

^{-1}

for

t \ge 0

. Determine the times when the particle is instantaneously at rest, and calculate the displacement from

t = 0

t = 3

Show worked answer →

At rest when $v = 0$ (M1): $3t^2 - 12t + 9 = 0$ , so $t^2 - 4t + 3 = 0$ and $(t - 1)(t - 3) = 0$ , giving $t = 1$ and $t = 3$ s (A1).

Displacement is $\int_0^3 v\,dt$ (M1): $\int_0^3 (3t^2 - 12t + 9)\,dt = [t^3 - 6t^2 + 9t]_0^3$ (A1).

Evaluate (M1): at $t = 3$ , $27 - 54 + 27 = 0$ ; at $t = 0$ , $0$ . So the displacement is $0 - 0 = 0$ m (A1).

Markers reward solving $v = 0$ , integrating, and evaluating between the limits.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)