Motion is described in physics using five key quantities: distance, displacement, speed, velocity, and acceleration. Graphs are the most powerful tool for analysing motion — the slope tells you velocity, and the area tells you displacement.
Describing Motion — Definitions and Graphs
Definition
Distance
Distance is the total length of the path travelled, regardless of direction; a scalar quantity. SI unit: metre (m).
Definition
Displacement (Δx)
Displacement is the change in position of an object; the straight-line distance from start to finish with direction; a vector quantity. SI unit: metre (m).
Definition
Speed
Speed is the rate of change of distance (distance ÷ time); a scalar quantity. SI unit: m·s⁻¹.
Definition
Velocity (v)
Velocity is the rate of change of displacement; a vector quantity. v = Δx/Δt. SI unit: m·s⁻¹.
Definition
Acceleration (a)
Acceleration is the rate of change of velocity; a = Δv/Δt; a vector quantity. SI unit: m·s⁻².
Definition
Uniform velocity
Uniform velocity is motion in which an object covers equal displacements in equal time intervals (constant velocity).
Definition
Uniform acceleration
Uniform acceleration is motion in which velocity changes by equal amounts in equal time intervals.
Formula
Average velocity
v_av = average velocity (m·s⁻¹), Δx = displacement (m), Δt = time interval (s)
SI unit: m·s⁻¹
Formula
Acceleration
a = acceleration (m·s⁻²), Δv = change in velocity (m·s⁻¹), Δt = time (s)
SI unit: m·s⁻²
Position-time vs Velocity-time Graphs
| Property | Position-time (x-t) | Velocity-time (v-t) |
|---|---|---|
| Slope gives | Velocity | Acceleration |
| Area gives | — (not applicable) | Displacement |
| Horizontal line | Object at rest | Constant velocity (zero acceleration) |
| Straight sloped line | Constant velocity | Constant acceleration |
| Curved line | Changing velocity (acceleration present) | Changing acceleration |
Worked Example
An object moves 60 m North in 12 s at constant velocity. Calculate its velocity and describe its x-t graph.
Given
- Δx = 60 m North
- Δt = 12 s
Find
v and shape of x-t graph
Solution
- 1v = Δx/Δt = 60/12 = 5 m·s⁻¹ North
- 2On the x-t graph: a straight line with positive slope (slope = 5 m·s⁻¹)
Worked Example
A car speeds up from 10 m·s⁻¹ to 30 m·s⁻¹ in 4 s. Calculate: (a) acceleration, (b) what the v-t graph looks like.
Given
- vᵢ = 10 m·s⁻¹
- vf = 30 m·s⁻¹
- Δt = 4 s
Find
(a) acceleration (b) v-t graph description
Solution
- 1(a) a = Δv/Δt = (30 − 10)/4 = 20/4 = 5 m·s⁻²
- 2(b) The v-t graph is a straight line starting at v = 10 m·s⁻¹ and ending at v = 30 m·s⁻¹; slope = 5 m·s⁻²
Exam Tip
To find displacement from a v-t graph: calculate the AREA under the graph (including sign). A rectangle = base × height. A triangle = ½ × base × height. Areas below the time axis represent negative displacement (opposite direction).
Practice Question
From a v-t graph, a car travels at constant 20 m·s⁻¹ for 8 s, then decelerates uniformly to rest in 4 s. Calculate the total displacement.
(5 marks)
Practice Question
On an x-t graph, point A is at x = 2 m at t = 0 s and point B is at x = 14 m at t = 4 s. (a) Calculate the average velocity. (b) What shape is the graph between A and B if velocity is constant?
(4 marks)
Instantaneous Velocity and Describing Motion Completely
Definition
Instantaneous velocity
Instantaneous velocity is the velocity of an object at a specific instant in time; equals the gradient of the tangent to the position-time graph at that point.
Definition
Average velocity
Average velocity is the total displacement divided by the total time taken for the journey.
On a curved x-t graph (where velocity is changing), the instantaneous velocity at any point equals the slope of the tangent drawn to the curve at that point. On a straight-line x-t graph, instantaneous velocity equals average velocity at every point on that segment — they are the same because the slope is constant.
Watch Out
A common mistake: confusing DISTANCE with DISPLACEMENT. Distance is the total path length (always positive, a scalar). Displacement is the straight-line change in position (can be negative if you moved backward, a vector). These are only equal when motion is in a single direction without reversing.
Worked Example
An object starts at x = 0 and moves to x = +20 m in 5 s, then returns to x = +8 m in the next 3 s. Calculate: (a) total distance, (b) displacement from start, (c) average speed, (d) average velocity for the whole trip.
Given
- Forward: +20 m in 5 s
- Backward: moves from x = 20 m back to x = 8 m (12 m backward) in 3 s
- Total time = 5 + 3 = 8 s
Find
(a) distance (b) displacement (c) average speed (d) average velocity
Solution
- 1(a) Total distance = 20 + 12 = 32 m
- 2(b) Displacement = final position − initial = +8 − 0 = +8 m
- 3(c) Average speed = total distance / total time = 32/8 = 4 m·s⁻¹
- 4(d) Average velocity = displacement / total time = +8/8 = +1 m·s⁻¹ (positive direction)
Practice Question
A cyclist travels 30 m East in 6 s, stops for 2 s, then travels 10 m West in 4 s. Calculate: (a) total distance, (b) resultant displacement, (c) average speed, (d) average velocity.
(6 marks)