Distance, displacement, speed, velocity and acceleration: WASSCE Physics

You sit still in a poda-poda while it races along Wilkinson Road. Are you moving? Relative to the vehicle, no; relative to the road, yes. Physics describes motion carefully, with five quantities you must never mix up: distance, displacement, speed, velocity and acceleration.

Learning objectives

By the end of this lesson you should be able to:

describe motion relative to a reference point, using a sign for direction
distinguish distance from displacement and speed from velocity
calculate average speed, average velocity and acceleration
convert between km h⁻¹ and m s⁻¹

Describing motion

A body is in motion when its position changes with time relative to a chosen reference point. For motion along a straight line, choose one direction as positive (say east). Then:

$+12$ m means 12 m east of the reference point;
$-5$ m means 5 m west of it.

The sign tells you the direction; the number tells you the size.

Distance and displacement

Definition

Distance is the total length of the path travelled. It is a scalar: it has size only.
Displacement is the change in position, measured in a straight line from start to finish in a stated direction. It is a vector.

A straight east-west road: a student walks 300 m east from the start to a shop, then 100 m west to the finish. The distance is 400 m; the displacement is 200 m east

Figure 1. Distance and displacement on a straight road. WASSCE Tutor diagram.

Worked example

A student walks 300 m east to a shop, then 100 m west. Find the distance travelled and the displacement.

Distance $= 300 + 100 = 400$ m.

Taking east as positive: displacement $= +300 - 100 = +200$ m, that is 200 m east of the start.

Distance can never be negative. Displacement can be positive, negative or zero, and its size can never be larger than the distance.

Worked example

A runner completes one 400 m lap of the National Stadium track and stops where she started. Find the distance and the displacement.

Distance $= 400$ m. She ends where she began, so displacement $= 0$ .

Do not say she "travelled 0 m": zero is the displacement, not the distance.

Speed and velocity

\text{average speed} = \dfrac{\text{total distance}}{\text{total time}} \qquad \text{average velocity} = \dfrac{\text{displacement}}{\text{time taken}}

\begin{gathered}\begin{aligned}&\text{average speed} \\ &= \dfrac{\text{total distance}}{\text{total time}}\end{aligned} \\[6pt] \begin{aligned}&\text{average velocity} \\ &= \dfrac{\text{displacement}}{\text{time taken}}\end{aligned}\end{gathered}

Speed is a scalar; velocity is a vector, so it carries a direction (or a sign). Both are measured in m s⁻¹.

Worked example

A cyclist rides 600 m east in 120 s, then 200 m west in 40 s. Find her average speed and average velocity for the whole journey.

Total distance $= 600 + 200 = 800$ m; total time $= 120 + 40 = 160$ s.

\text{average speed} = \dfrac{800}{160} = 5.0\ \text{m s}^{-1}

\begin{aligned}&\text{average speed} \\ &= \dfrac{800}{160} \\ &= 5.0\ \text{m s}^{-1}\end{aligned}

Displacement $= +600 - 200 = +400$ m (east).

\text{average velocity} = \dfrac{400}{160} = 2.5\ \text{m s}^{-1}\ \text{east}

\begin{aligned}&\text{average velocity} \\ &= \dfrac{400}{160} \\ &= 2.5\ \text{m s}^{-1}\ \text{east}\end{aligned}

The two answers differ because she turned back.

Converting km h⁻¹ and m s⁻¹

1\ \text{km h}^{-1} = \dfrac{1000\ \text{m}}{3600\ \text{s}} = \dfrac{5}{18}\ \text{m s}^{-1}

\begin{aligned}&1\ \text{km h}^{-1} \\ &= \dfrac{1000\ \text{m}}{3600\ \text{s}} \\ &= \dfrac{5}{18}\ \text{m s}^{-1}\end{aligned}

km h⁻¹ → m s⁻¹: multiply by $\dfrac{5}{18}$ . So $72\ \text{km h}^{-1} = 72 \times \dfrac{5}{18} = 20\ \text{m s}^{-1}$ .
m s⁻¹ → km h⁻¹: multiply by $\dfrac{18}{5}$ . So $15\ \text{m s}^{-1} = 54\ \text{km h}^{-1}$ .

Worked example

A vehicle covers 1.8 km in 3 minutes. Find its average speed in m s⁻¹ and in km h⁻¹.

v = \dfrac{1800\ \text{m}}{180\ \text{s}} = 10\ \text{m s}^{-1} = 10 \times \dfrac{18}{5} = 36\ \text{km h}^{-1}

\begin{aligned}v &= \dfrac{1800\ \text{m}}{180\ \text{s}} \\ &= 10\ \text{m s}^{-1} \\ &= 10 \times \dfrac{18}{5} \\ &= 36\ \text{km h}^{-1}\end{aligned}

Instantaneous speed

Average speed describes a whole journey. Instantaneous speed is the speed at one moment: the number on a car's speedometer. On a displacement–time graph it is the gradient of the tangent at that instant (see the next lesson).

Acceleration

Definition

Acceleration is the rate of change of velocity:

a = \dfrac{v - u}{t}

where $u$ is the initial velocity, $v$ the final velocity and $t$ the time taken. Its unit is m s⁻². Acceleration is a vector.

Worked example

A bus speeds up from 5 m s⁻¹ to 17 m s⁻¹ in 6 s. Find its acceleration.

a = \dfrac{17 - 5}{6} = 2.0\ \text{m s}^{-2}

\begin{aligned}a &= \dfrac{17 - 5}{6} \\ &= 2.0\ \text{m s}^{-2}\end{aligned}

When a body slows down, its acceleration is opposite to its velocity. This is called deceleration or retardation.

Worked example

A car moving east at 20 m s⁻¹ stops uniformly in 5 s. Taking east as positive, find its acceleration.

a = \dfrac{0 - 20}{5} = -4.0\ \text{m s}^{-2}

\begin{aligned}a &= \dfrac{0 - 20}{5} \\ &= -4.0\ \text{m s}^{-2}\end{aligned}

The minus sign means the acceleration points west, against the motion. The car has a deceleration of 4.0 m s⁻².

Common mistake

"Negative acceleration" does not always mean slowing down. A body moving west (negative) that speeds up also has a negative acceleration. Decide the sign from your chosen positive direction, and say in words whether the body is speeding up or slowing down.

In Sierra Leone

Okada riders in Bo and Makeni know acceleration by feel. Pulling away from a junction, a bike might reach 12 m s⁻¹ (about 43 km h⁻¹) in 6 s, an acceleration of 2 m s⁻². Braking hard from the same speed to rest in 2 s is a deceleration of 6 m s⁻², which is why passengers lurch forward.

Key points

Choose a positive direction; signs show direction along a line.
Distance (scalar) is the path length; displacement (vector) is the straight-line change in position, with direction.
Average speed = total distance ÷ time; average velocity = displacement ÷ time.
km h⁻¹ × 5/18 = m s⁻¹; m s⁻¹ × 18/5 = km h⁻¹.
$a = (v - u)/t$ , in m s⁻²; deceleration is acceleration opposite to the motion.

WASSCE examination focus

WAEC tests this

WASSCE asks you to define distance, displacement, speed, velocity and acceleration and to classify them as scalars or vectors. Calculations involve journeys with a change of direction (where average speed and average velocity differ), unit conversion between km h⁻¹ and m s⁻¹, and acceleration or deceleration from $a = (v - u)/t$ . Always state the direction of a vector answer.

Quick check

A student walks 250 m north and then 90 m south. Find the distance and the displacement.
Convert 72 km h⁻¹ to m s⁻¹.
A cyclist travels 500 m east in 100 s, then 200 m west in 50 s. Find the average speed and the average velocity.
A car's velocity rises from 8 m s⁻¹ to 26 m s⁻¹ in 6 s. Find its acceleration.
Is speed a scalar or a vector? Is velocity?

Answers: 1. Distance 340 m; displacement 160 m north. 2. 20 m s⁻¹. 3. Speed $= 700 \div 150 = 4.67$ m s⁻¹; velocity $= 300 \div 150 = 2.0$ m s⁻¹ east. 4. $(26 - 8) \div 6 = 3.0$ m s⁻². 5. Speed is a scalar; velocity is a vector.

Distance, displacement, speed, velocity and acceleration

Learning objectives

Describing motion

Distance and displacement

Speed and velocity

Converting km h⁻¹ and m s⁻¹

Instantaneous speed

Acceleration

Key points

WASSCE examination focus

Quick check

Keep going with motion in a straight line

More lessons in Motion in a Straight Line