TASMath MethodsSyllabus dot point

How do calculus operations connect position, velocity and acceleration in straight-line motion?

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

How to move between position, velocity and acceleration using differentiation and integration, with sign and direction interpretation for straight-line motion in TCE Mathematics Methods.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

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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.

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 $v=0$ .

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 $x(t)$ .

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.