Coaching Centre
Describing motion: scalars and vectors
Scalars and vectors, frames of reference, position, distance and displacement, speed and velocity, average speed and average velocity, with sign conventions for motion in a straight line.
At the Sydney 2000 Olympics, Cathy Freeman ran 400 m in 49.11 seconds and won gold. Her average speed was more than 8 metres per second. Her average velocity was exactly zero. Both statements are correct, and by the end of this page you will be able to explain why.
In Year 10 you met speed, distance and simple motion graphs. Year 11 Physics is far more careful with words. "Distance" and "displacement" are different quantities, so are "speed" and "velocity", and HSC markers take marks off when they are mixed up. This page sets up the language for the rest of the module.
Scalars and vectors
A scalar has size (magnitude) only: a number with a unit. A vector has size and direction. You can't fully describe a vector without saying which way it points.
| Scalars | Vectors |
|---|---|
| distance (m) | displacement (m) |
| speed (m s⁻¹) | velocity (m s⁻¹) |
| time (s) | acceleration (m s⁻²) |
| mass (kg) | force (N) |
| energy (J) | momentum (kg m s⁻¹) |
On paper, vectors are drawn as arrows. The length of the arrow shows the magnitude, and the arrow shows the direction. In print a vector is often written in bold ($\mathbf{v}$) or with an arrow ($\vec{v}$); in your own working, an arrow or a squiggle under the letter is fine.
Position and frames of reference
Every description of motion is made relative to something: a frame of reference. Usually that is the ground, but it doesn't have to be. A passenger sitting in a train is at rest relative to the train and moving at 100 km/h relative to the platform. Neither description is wrong; they use different frames. You'll use this idea properly in 1H.
For motion in a straight line, choose:
- an origin, the zero point you measure from, and
- a positive direction. The opposite direction is negative.
Then position $x$ is a signed number: $x = -300$ m means 300 m on the negative side of the origin. Once you have chosen a sign convention for a problem, stick to it for every vector in that problem.
Distance and displacement
- Distance is the total length of the path travelled. It is a scalar and can never decrease.
- Displacement is the change in position: a straight-line vector from the start point to the end point. $\;s = \Delta x = x_{\text{final}} - x_{\text{initial}}$.
Displacement depends only on where you start and finish, not on the route. A journey that ends where it began always has zero displacement, however long it was.
Speed and velocity
Average speed and average velocity are defined differently, and that is where the marks are:
$$\text{average speed} = \frac{\text{distance travelled}}{\text{time taken}} \qquad v_{\text{avg}} = \frac{\Delta r}{\Delta t} = \frac{\text{displacement}}{\text{time taken}}$$The formula sheet writes displacement as $\Delta r$. In one dimension this is the same as $s$ or $\Delta x$. Average velocity is a vector: it has the direction of the displacement. Average speed is a scalar and is never negative.
Two more terms:
- Uniform (rectilinear) motion: moving in a straight line at constant velocity. Equal displacements in equal times; no acceleration.
- Uniformly accelerated motion: velocity changes by the same amount every second. A car pulling away steadily, or a ball in free fall (ignoring air resistance).
Units: the SI unit of velocity is m s⁻¹ (also written m/s). To convert km/h to m/s, divide by 3.6. To go from m/s to km/h, multiply by 3.6. That works because $1\ \text{km/h} = \frac{1000\ \text{m}}{3600\ \text{s}} = \frac{1}{3.6}$ m/s.
A delivery van drives 3.0 km east in 200 s, stops for 60 s, drives 1.2 km west in 100 s, stops for 40 s, then drives 4.2 km east in 210 s. Find (a) the distance travelled, (b) the displacement, (c) the average speed and (d) the average velocity for the whole trip.
- Choose Walk to the shop and back and press Play. At the end, the distance is 800 m but the displacement is 0 m. Check the average speed is 1.16 m/s and the average velocity is 0 m/s.
- Drag the time slider to the moment the walker reaches the shop. Why are the average speed and average velocity equal up to this point?
- Choose Delivery van and check the worked example above: 8400 m, +6000 m, 13.77 m/s and 9.84 m/s.
- Switch the positive direction to west. Which readouts change sign, and which don't change at all?
- In Build your own, find a journey where the average speed is twice the size of the average velocity.
Test yourself
Which list contains only vector quantities?
Speed, distance, time, mass and energy are scalars.
A cyclist rides 6.0 km north and then 2.0 km south in a total of 20 minutes. Their average velocity is
Displacement $= 6.0 - 2.0 = 4.0$ km north $= 4000$ m. $v_{\text{avg}} = 4000 \div 1200 = 3.3$ m s⁻¹ north. Option B is 8 km in a third of an hour, the average speed.
For any journey, which statement is always true?
Distance is at least as large as the size of the displacement, and both are divided by the same time. They are equal only for travel in one direction along a straight line.
A car's speed is 90 km/h. In SI units this is
$90 \div 3.6 = 25$ m s⁻¹.
Label the walk to the shop and back.
Sort these quantities.
A vector has both magnitude and . Displacement is the change in . Average velocity is displacement divided by . After one complete lap of a track, a runner's displacement is . To convert km/h to m/s, divide by .
A swimmer completes two laps of a 50 m pool (100 m in total) in 80 s, finishing where she started. Calculate her average speed and average velocity, and explain why the two answers are different.
Average speed = distance ÷ time = 100 ÷ 80 = 1.25 m s⁻¹.
She finishes where she started, so her displacement is 0 m and her average velocity is 0 ÷ 80 = 0 m s⁻¹.
The answers differ because average speed is based on distance, the total length of the path (a scalar), while average velocity is based on displacement, the straight-line change in position (a vector). Swimming out and back gives a large distance but no change in position.