Kinematics: position, velocity and acceleration (TCE Mathematics Methods, Tasmania)

Syllabus dot point

“Relate position, velocity and acceleration through differentiation and integration for rectilinear motion.”

8 min readUpdated 2 Jun 2026

Quick answer

For straight-line motion, velocity is the derivative of position, $v=\dfrac{dx}{dt}$ , and acceleration is the derivative of velocity, $a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}$ . Going the other way, integrating acceleration gives velocity and integrating velocity gives position, with constants fixed by initial conditions.

What this dot point is asking

Kinematics is the headline application of calculus in this course. It links the second derivative from Unit 3 with the integration of Unit 4, so a strong grasp here pays off twice.

The three quantities and how calculus connects them

Position $x(t)$ measures where an object is along a line at time $t$ . Differentiating once gives velocity, differentiating again gives acceleration.

Differentiation links

$v(t) = \dfrac{dx}{dt}$ (velocity is the rate of change of position)

$a(t) = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ (acceleration is the rate of change of velocity)

Reversing the chain, integration recovers the quantity above:

v(t) = \int a(t)\,dt, \qquad x(t) = \int v(t)\,dt.

Each integration introduces a constant, which you find from a given initial condition such as the velocity at $t=0$ or the starting position.

Reading the signs

Sign carries direction. A positive velocity means the object moves in the positive direction; a negative velocity means it moves backwards. The object is momentarily at rest when $v=0$ .

Speeding up versus slowing down

Speed is the magnitude of velocity, $|v|$ . An object speeds up when velocity and acceleration have the same sign, and slows down when they have opposite signs. So a negative velocity with a positive acceleration means the object is slowing down, not speeding up.

Distance versus displacement

Displacement is the change in position, $x(t_2)-x(t_1)$ , and can be negative. Distance travelled is the total length of the path and is never negative. If the object reverses direction within the interval (where $v=0$ and changes sign), you must split the journey at that instant and add the magnitudes of each segment.

Finding velocity, rest time and acceleration

Problem

A particle moves so that its position is $x(t) = t^3 - 6t^2 + 9t$ metres, with $t$ in seconds and $t\ge 0$ . Find its velocity, the times when it is at rest, and its acceleration at $t=1$ .

Solution

Velocity is the derivative:

v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9 = 3(t-1)(t-3).

The particle is at rest when $v=0$ , that is at $t=1$ s and $t=3$ s.

Acceleration is the second derivative:

a(t) = \frac{dv}{dt} = 6t - 12.

At $t=1$ : $a(1) = 6(1) - 12 = -6 \text{ m/s}^2$ . Since $v(1)=0$ and the velocity changes sign here, the particle is reversing direction at this instant.

Recovering motion from acceleration

A common second style of question gives the acceleration and an initial condition and asks you to integrate. For example, if $a(t) = 6t$ with $v(0)=2$ , then $v(t) = \int 6t\,dt = 3t^2 + C$ , and $v(0)=2$ gives $C=2$ , so $v(t) = 3t^2 + 2$ . Integrating again with a starting position recovers $x(t)$ .

Recovering position from acceleration

A particle has acceleration $a(t) = 6t - 4$ m/s $^{2}$ . At $t = 0$ its velocity is $5$ m/s and its position is $2$ m. Find $v(t)$ and $x(t)$ .

Step 1: Integrate acceleration to find velocity

Since velocity is the antiderivative of acceleration, integrate $a(t)$ term by term. Each integration introduces an unknown constant $C$ , which the initial condition will determine:

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

Step 2: Use the initial velocity to find $C$

Substitute the known condition $v(0) = 5$ into the expression above. Every term with $t$ vanishes, leaving $C$ alone:

v(0) = 3(0)^{2} - 4(0) + C = C = 5,

so $v(t) = 3t^{2} - 4t + 5$ .

Step 3: Integrate velocity to find position

Integrate $v(t)$ to recover position. A second constant of integration $D$ appears because this is another independent antidifferentiation:

x(t) = \int (3t^{2} - 4t + 5)\,dt = t^{3} - 2t^{2} + 5t + D.

Step 4: Use the initial position to find $D$

Substitute $x(0) = 2$ to pin down $D$ :

x(0) = 0 - 0 + 0 + D = D = 2,

so $x(t) = t^{3} - 2t^{2} + 5t + 2$ .

Final answer: $v(t) = 3t^{2} - 4t + 5$ m/s and $x(t) = t^{3} - 2t^{2} + 5t + 2$ m. Each integration introduces one constant, and each initial condition fixes exactly one of them.

Using a velocity-time graph

A velocity-time graph encodes everything at a glance. The gradient of the graph at any instant is the acceleration, and the signed area between the graph and the time axis over an interval is the displacement. Area below the axis (negative velocity) counts as negative displacement, so for total distance travelled you add the magnitudes of the areas above and below separately. This graphical reading is the geometric counterpart of the calculus: gradient corresponds to differentiation, area to integration.

Confusing displacement with distance travelled

If asked for distance travelled over an interval where the particle changes direction, integrating velocity directly gives displacement, which can understate the true distance because forward and backward motion cancel. Find where $v=0$ , split the interval there, evaluate each piece, and add the absolute values to get the genuine distance.

Summary

Differentiate position to get velocity and velocity to get acceleration; integrate to go back the other way, fixing each constant with an initial condition. Use the sign of velocity for direction and the matching of velocity and acceleration signs to decide whether the object is speeding up or slowing down. Always split at the rest points when a question asks for distance travelled rather than displacement.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Calculator-free. A particle moves in a straight line so that its position is

x(t) = t^{3} - 9t^{2} + 24t

metres, with

t

in seconds and

t \ge 0

. (a) Find the velocity

v(t)

and the times at which the particle is at rest. (b) Find the acceleration when

t = 2

. (c) Determine the total distance travelled in the first

5

seconds.

Show worked answer →

(a) Differentiate: $v(t) = \dfrac{dx}{dt} = 3t^{2} - 18t + 24 = 3(t - 2)(t - 4)$ . The particle is at rest when $v = 0$ , i.e. at $t = 2$ s and $t = 4$ s.

(b) Differentiate again: $a(t) = \dfrac{dv}{dt} = 6t - 18$ , so $a(2) = 6(2) - 18 = -6$ m/s $^{2}$ .

(c) The particle reverses at $t = 2$ and $t = 4$ , so split $[0, 5]$ there. Positions: $x(0) = 0$ , $x(2) = 8 - 36 + 48 = 20$ , $x(4) = 64 - 144 + 96 = 16$ , $x(5) = 125 - 225 + 120 = 20$ .
Distances of each leg: $|20 - 0| + |16 - 20| + |20 - 16| = 20 + 4 + 4 = 28$ m.
One mark for $v(t)$ , one for the rest times, one for $a(2)$ , two for splitting at the rest points and summing the magnitudes.

Practice4 marks

Calculator-assumed. A body starts from rest at the origin with acceleration

a(t) = 12 - 6t

m/s

^{2}

. (a) Find the velocity

v(t)

. (b) Find the position

x(t)

. (c) Find the maximum velocity and the time at which it occurs.

Show worked answer →

(a) Integrate acceleration: $v(t) = \displaystyle\int (12 - 6t)\,dt = 12t - 3t^{2} + C$ . At rest at $t = 0$ gives $v(0) = 0$ , so $C = 0$ and $v(t) = 12t - 3t^{2}$ .

(b) Integrate velocity: $x(t) = \displaystyle\int (12t - 3t^{2})\,dt = 6t^{2} - t^{3} + D$ . Starting at the origin gives $x(0) = 0$ , so $D = 0$ and $x(t) = 6t^{2} - t^{3}$ .

(c) Maximum velocity occurs when $a = v'(t) = 0$ : $12 - 6t = 0$ , so $t = 2$ s. Then $v(2) = 24 - 12 = 12$ m/s. Markers reward both constants of integration found from the initial conditions and the correct maximum.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC

What this dot point is asking

The three quantities and how calculus connects them

Reading the signs

Distance versus displacement

Problem

Solution

Recovering motion from acceleration

Step 1: Integrate acceleration to find velocity

Step 2: Use the initial velocity to find CCC

Step 3: Integrate velocity to find position

Step 4: Use the initial position to find DDD

Using a velocity-time graph

Summary

Exam-style questions

Practise this

Sources & how we know this

Step 2: Use the initial velocity to find $C$

Step 4: Use the initial position to find $D$