ScotlandMathematics of MechanicsSyllabus dot point

How do you describe straight-line motion, and how do calculus and the constant-acceleration equations connect displacement, velocity and acceleration?

Work with rectilinear motion: relate displacement, velocity and acceleration by differentiation and integration, use the equations of motion for constant acceleration, and interpret motion-time graphs.

A focused answer to the SQA Advanced Higher Mathematics of Mechanics content on rectilinear motion, linking displacement, velocity and acceleration through differentiation and integration, applying the constant-acceleration equations, and reading velocity-time and displacement-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
Displacement, velocity and acceleration by calculus
The equations of motion for constant acceleration
Motion-time graphs
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What this dot point is asking

Mechanics begins with kinematics: describing how something moves without yet asking why. For motion along a straight line the SQA wants you to move fluently between displacement, velocity and acceleration, both with calculus (when acceleration varies) and with the constant-acceleration equations (when it does not), and to read the same information off a motion-time graph.

Displacement, velocity and acceleration by calculus

The three quantities form a chain linked by differentiation in one direction and integration in the other. Displacement $s$ is a signed position relative to a fixed origin; velocity $v$ is its rate of change; acceleration $a$ is the rate of change of velocity.

This is the route to take whenever acceleration is not constant, for example when $a$ is given as a function of $t$ . Each integration introduces an arbitrary constant that you fix using a known value (an initial velocity or position).

Variable acceleration: integrating for velocity and displacement

A particle has acceleration $a = 6t - 4$ m s $^{-2}$ , with $v = 1$ m s $^{-1}$ and $s = 0$ at $t = 0$ .

Step 1: Integrate the acceleration to find velocity

Because the acceleration varies with time, the suvat equations cannot be used. Integrate $a$ with respect to $t$ , introducing a constant of integration $c$ , then use the initial velocity to fix it:

v = \int (6t - 4)\,dt = 3t^2 - 4t + c.

With $v = 1$ at $t = 0$ : $c = 1$ , so $v = 3t^2 - 4t + 1$ .

Step 2: Integrate the velocity to find displacement

Integrate $v$ again, fixing the new constant from the initial position:

s = \int (3t^2 - 4t + 1)\,dt = t^3 - 2t^2 + t + k.

With $s = 0$ at $t = 0$ : $k = 0$ , so $s = t^3 - 2t^2 + t$ .

Step 3: Evaluate at a specific time

At $t = 2$ , substitute into both expressions:

v = 12 - 8 + 1 = 5\ \text{m s}^{-1}, \quad s = 8 - 8 + 2 = 2\ \text{m.}

Final answer: at $t = 2\ \text{s}$ the particle has velocity $5\ \text{m s}^{-1}$ and displacement $2\ \text{m}$ .

The equations of motion for constant acceleration

When acceleration is constant, integrating gives the familiar suvat equations. Each uses four of the five quantities $s, u, v, a, t$ , so you choose the one that contains the three you know and the one you want.

These are not separate facts to memorise blindly: every one follows from $a = \dfrac{dv}{dt}$ being constant. The third, $v^2 = u^2 + 2as$ , is the time-free equation and is usually the fastest when time is neither given nor wanted.

Decelerating train: acceleration and stopping distance

A train decelerates uniformly from $30$ m s $^{-1}$ and stops in $12$ s.

Step 1: List the known quantities and choose the right equation

The acceleration is constant (suvat applies), and we know $u = 30$ , $v = 0$ , $t = 12$ . We want $a$ first. The equation $v = u + at$ uses exactly these four quantities, so pick it:

0 = 30 + 12a, \quad a = -2.5\ \text{m s}^{-2}.

The negative sign confirms deceleration.

Step 2: Find the stopping distance

Now find $s$ without using $a$ (to avoid carrying rounding errors). The average-velocity equation $s = \tfrac{1}{2}(u + v)t$ needs only the speeds and the time:

s = \tfrac{1}{2}(30 + 0)(12) = 180\ \text{m.}

Final answer: the deceleration is $2.5\ \text{m s}^{-2}$ and the stopping distance is $180\ \text{m}$ .

Motion-time graphs

A graph carries the same information as the calculus. On a displacement-time graph the gradient at a point is the velocity. On a velocity-time graph the gradient is the acceleration and, crucially, the area between the graph and the time axis is the displacement.

Because area under a velocity-time graph is $\int v\,dt$ , finding distance from a graph and finding it by integration are the same calculation. For a graph made of straight segments you can use areas of triangles and trapezia instead of integrating.

Try this

Q1. A particle moves with $v = 4t - t^2$ m s $^{-1}$ . Find its acceleration at $t = 3$ s. [2 marks]

Cue. $a = \dfrac{dv}{dt} = 4 - 2t$ , so at $t = 3$ , $a = 4 - 6 = -2$ m s $^{-2}$ .

Q2. A stone is thrown so that it travels $45$ m while accelerating uniformly from rest in $3$ s. Find the acceleration. [2 marks]

Cue. $s = ut + \tfrac{1}{2}at^2$ with $u = 0$ : $45 = \tfrac{1}{2}a(9)$ , so $a = 10$ m s $^{-2}$ .

Exam-style practice questions

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

AH style: calculus kinematics5 marksA particle moves in a straight line so that its velocity is

v = 3t^2 - 12t + 9

m s

^{-1}

at time

t

seconds. Find the times when the particle is instantaneously at rest, and the total distance travelled in the first 3 seconds.

Show worked answer →

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

Displacement is $s = \int v\,dt = t^3 - 6t^2 + 9t$ (taking $s = 0$ at $t = 0$ ) (1 mark). The velocity changes sign at $t = 1$ , so distance must be found piecewise.

From $t = 0$ to $1$ : $s(1) - s(0) = (1 - 6 + 9) - 0 = 4$ m. From $t = 1$ to $3$ : $s(3) - s(1) = (27 - 54 + 27) - 4 = -4$ m, so $4$ m travelled back (2 marks).

Total distance $= 4 + 4 = 8$ m (1 mark). Markers reward integrating for $s$ , splitting at the turning point, and adding the magnitudes rather than the signed displacements.

AH style: constant acceleration4 marksA car accelerates uniformly from

8

m s

^{-1}

20

m s

^{-1}

while travelling

84

m. Find its acceleration and the time taken.

Show worked answer →

Use $v^2 = u^2 + 2as$ with $u = 8$ , $v = 20$ , $s = 84$ : $400 = 64 + 168a$ , so $a = \dfrac{336}{168} = 2$ m s $^{-2}$ (2 marks).

Then $v = u + at$ gives $20 = 8 + 2t$ , so $t = 6$ s (2 marks). Markers reward choosing the equation that avoids the unknown time first, then a second equation for $t$ ; the alternative $s = \tfrac{1}{2}(u + v)t$ giving $84 = 14t$ also earns full marks.

Related dot points

Sources & how we know this

Advanced Higher Mathematics of Mechanics Course Specification (C802 77) — SQA (2019)