Displacement, velocity and acceleration (QCE Physics Unit 2)

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).

Examples in context

Example 1. A Cairns light-rail train accelerates from rest over $400 \text{ m}$ between Smithfield stops, reaching $14 \text{ m s}^{-1}$ in $60 \text{ s}$. Average velocity is $\bar{v} = \Delta x / \Delta t = 400/60 = 6.67 \text{ m s}^{-1}$, distinct from the instantaneous velocity at the end of the leg. Average acceleration is $\bar{a} = \Delta v / \Delta t = 14/60 = 0.23 \text{ m s}^{-2}$. The QCAA Unit 2 distinction between average and instantaneous quantities is exactly the data-test analysis required when track sensors stream tachograph data into a stimulus stem.

Example 2. A Sunshine Coast tidal-study buoy logs displacement of a drifter every $30 \text{ s}$. Over one hour the drifter moves $720 \text{ m}$ along a curved track but its net displacement is only $480 \text{ m}$ east. The distance (scalar) is therefore $720 \text{ m}$, the displacement (vector) is $480 \text{ m}$ east, and average speed ($0.20 \text{ m s}^{-1}$) exceeds the magnitude of average velocity ($0.133 \text{ m s}^{-1}$). QCAA Unit 2 IA1 stimulus often uses such an oceanographic dataset to test the scalar versus vector distinction.

Try this

Q1. Define displacement, velocity and acceleration and identify which are vectors. [3 marks]

• Cue. Displacement (vector) = change in position; velocity (vector) = rate of change of displacement; acceleration (vector) = rate of change of velocity.

Q2. A car covers $1200 \text{ m}$ around a loop in $80 \text{ s}$, returning to its start. State the distance, the displacement, the average speed and the average velocity. [3 marks]

• Cue. Distance $1200 \text{ m}$; displacement $0$; average speed $15 \text{ m s}^{-1}$; average velocity $0$.

Q3. A Cairns light-rail train accelerates from rest to $14 \text{ m s}^{-1}$ in $20 \text{ s}$. (a) Calculate the average acceleration. (b) Sketch a velocity-time graph and identify the area-under-graph as displacement. (c) Explain why instantaneous velocity at $t = 10 \text{ s}$ may differ from the average if acceleration is non-uniform. [2+3+2 marks; ISMG: Analysis and interpretation]

• Cue. (a) $0.70 \text{ m s}^{-2}$; (b) area = displacement, $140 \text{ m}$ if uniform; (c) instantaneous = tangent slope, may be greater or less than chord.