Recall, describe and apply the concepts of position, displacement, distance, speed, velocity and acceleration, distinguishing between scalar and vector quantities and between average and instantaneous values

What this dot point is asking

QCAA expects you to use the standard kinematic vocabulary with precision. The same SI units appear in pairs (distance and displacement in metres; speed and velocity in m s$^{-1}$) but the pair members are not interchangeable. The dot point also requires the distinction between average values (over an interval) and instantaneous values (at one moment).

Definitions

Position. The location of an object relative to a reference point, given as a coordinate (often $x$ or $\vec{r}$). Vector.

Distance. The total length of the path travelled. Scalar.

Displacement. The change in position from start to finish, $\Delta \vec{r} = \vec{r}_f - \vec{r}_i$. Vector. Independent of path.

Speed. Distance per unit time. Scalar.

Velocity. Rate of change of displacement. Vector.

Acceleration. Rate of change of velocity. Vector.

Acceleration has SI unit m s$^{-2}$ and points in the direction of the change in velocity, not the direction of motion. A car slowing down has acceleration opposite its velocity.

Average vs instantaneous

Average quantities use the endpoints of an interval. Instantaneous quantities use the limit as $\Delta t \to 0$, equivalent to the slope of the position-time graph (instantaneous velocity) or the velocity-time graph (instantaneous acceleration) at that point.

A speedometer reads instantaneous speed. A police speed trap measuring time over a known distance reads average speed.

Sign conventions

Pick a positive direction at the start of a problem and apply it consistently to position, velocity and acceleration. A negative velocity means motion in the negative direction; a negative acceleration means the velocity is becoming more negative (which can mean speeding up if velocity is already negative).

Worked example

A car has position $x(t) = 2.0t^2 - 4.0t + 1.0$ (metres, with $t$ in seconds). Find the velocity and acceleration at $t = 3.0$ s.

Velocity: $v = \frac{dx}{dt} = 4.0 t - 4.0$.

At $t = 3.0$ s: $v = 4.0(3.0) - 4.0 = 8.0$ m s$^{-1}$.

Acceleration: $a = \frac{dv}{dt} = 4.0$ m s$^{-2}$ (constant, so instantaneous = average).

Common traps

Treating speed and velocity as identical. A car driving in a circle at constant speed has zero average velocity over a full lap (start position equals end position).

Forgetting direction on vectors. QCAA penalises numerical answers for displacement, velocity and acceleration that omit direction.

Confusing slowing down with negative acceleration. A negative acceleration only means slowing down if the velocity is positive. If velocity is also negative, the object is speeding up.

Using an average value instantaneously. Average velocity over $10$ s is not the velocity at $t = 5$ s unless the motion is uniformly accelerated.

In one sentence

Distance and speed are scalar (path-length and rate of motion); displacement, velocity and acceleration are vectors (change in position, rate of change of displacement, rate of change of velocity), with average values over an interval and instantaneous values from the slope of a position-time or velocity-time graph.