Language of Kinematics

Understanding the fundamental quantities that describe motion: displacement, velocity, and acceleration

SPEC 7.1 • KINEMATICS

Scalar vs Vector Quantities

In mechanics, we must distinguish between two types of physical quantities:

Scalar Quantities

Magnitude only

Distance

Speed

Time

Mass

Energy

Vector Quantities

Magnitude AND direction

Displacement

Velocity

Acceleration

Force

Momentum

Key Concept

The distinction matters because vectors can be negative (indicating opposite direction), while scalars are always positive or zero.

Distance and Displacement

Distance

Distance is the total length of the path travelled. It is a scalar quantity — always positive.

Displacement

Displacement is the change in position from start to finish. It is a vector quantity — can be positive, negative, or zero.

Important

Distance and displacement are NOT the same! If you walk 5 m east then 5 m west, your distance is 10 m but your displacement is 0 m.

Speed and Velocity

Speed

Speed is the rate of change of distance with respect to time. It is a scalar.

Average Speed = \frac{Total Distance}{Total Time}

Velocity

Velocity is the rate of change of displacement with respect to time. It is a vector.

Average Velocity = \frac{Displacement}{Time} = \frac{Δ s}{Δ t}

Tip

Speed is the magnitude of velocity. If velocity is −5 m/s, the speed is 5 m/s.

Acceleration

Definition

Acceleration is the rate of change of velocity with respect to time. It is a vector quantity.

a = \frac{Δ v}{Δ t} = \frac{v - u}{t}

Understanding the signs of velocity and acceleration:

VelocityAccelerationMotion

Positive (+)Positive (+)Speeding up in positive direction

Positive (+)Negative (−)Slowing down in positive direction

Negative (−)Positive (+)Slowing down in negative direction

Negative (−)Negative (−)Speeding up in negative direction

Units and Notation

Standard SI units for kinematic quantities:

QuantitySymbolSI Unit

Displacement

s

or

r

metres (m)

Velocity

v

metres per second (m/s or m s⁻¹)

Acceleration

a

metres per second squared (m/s² or m s⁻²)

Time

t

seconds (s)

Sign Convention

Always define a positive direction at the start of a problem. Quantities in the opposite direction are negative.

Relationships Between Quantities

The kinematic quantities are connected through calculus:

Key Relationships

v = \frac{d s}{d t}

Velocity is the derivative of displacement

a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}

Acceleration is the derivative of velocity

Tip

This means: integrate acceleration to get velocity, integrate velocity to get displacement. We'll explore this fully in Lesson 7.06.

Worked Example01

Classify each of the following as scalar or vector:

(a) 50 km/h (b) 30 m due north (c) 10 s (d) −5 m/s² (e) 25 kg

a)

50 km/h — This is a speed (no direction given) → Scalar

b)

30 m due north — This is displacement (has direction) → Vector

c)

10 s — This is time → Scalar

d)

−5 m/s² — This is acceleration (the negative indicates direction) → Vector

e)

25 kg — This is mass → Scalar

Worked Example02

A cyclist travels 12 km in 30 minutes. Calculate the average speed in m/s.

1

Convert units: Distance = 12 km = 12,000 m

2

Convert units: Time = 30 min = 30 × 60 = 1800 s

3

Apply formula: Average Speed = Distance ÷ Time

4

Average Speed = 12000 ÷ 1800 = 6.67 m/s (3 s.f.)

Answer: 6.67 m/s

Worked Example03

A particle moves along a straight line. It travels 8 m to the right, then 3 m to the left. Taking right as positive, find:

(a) the total distance travelled

(b) the displacement

a)

Total distance = 8 + 3 = 11 m

•

(Distance is the total path length — always positive)

b)

Displacement = +8 + (−3) = +8 − 3 = +5 m

•

(Or 5 m to the right)

Answers: (a) 11 m (b) 5 m (or +5 m)

Worked Example04

A runner completes one lap of a 400 m circular track in 50 seconds, returning to the start. Calculate:

(a) the average speed

(b) the average velocity

a)

Average speed = Total distance ÷ Time

•

Average speed = 400 ÷ 50 = 8 m/s

b)

Displacement = 0 m (returns to start)

•

Average velocity = Displacement ÷ Time = 0 ÷ 50 = 0 m/s

Answers: (a) 8 m/s (b) 0 m/s

Worked Example05

A particle has velocity −4 m/s and acceleration +2 m/s². Describe its motion. What happens to its speed?

1

Velocity is negative: the particle is moving in the negative direction

2

Acceleration is positive: the velocity is increasing (becoming less negative)

3

Since v and a have opposite signs, the particle is slowing down

4

The speed (magnitude of velocity) is decreasing

Answer: The particle is moving in the negative direction and slowing down. Its speed is decreasing.

Worked Example06

Convert a velocity of 72 km/h to m/s.

1

72 km/h means 72 kilometres per hour

2

Convert km to m: 72 km = 72 × 1000 = 72,000 m

3

Convert hours to seconds: 1 hour = 60 × 60 = 3600 s

4

Velocity = 72,000 ÷ 3600 = 20 m/s

Answer: 20 m/s

To convert km/h to m/s: divide by 3.6

To convert m/s to km/h: multiply by 3.6

Worked Example07

An object moves with velocity 15 m/s at an angle of 30° above the horizontal. Express the velocity in terms of unit vectors i (horizontal) and j (vertical).

1

Horizontal component:

v_{x} = 15 cos 30° = 15 \times \frac{3}{2} = \frac{15 3}{2} \approx 13.0

m/s

2

Vertical component:

v_{y} = 15 sin 30° = 15 \times \frac{1}{2} = 7.5

m/s

3

In vector form:

v = v_{x} i + v_{y} j

Answer: $v = \frac{15 3}{2} i + 7.5 j$ m/s or $v = (13.0 i + 7.5 j)$ m/s

Practice

Question 1: A car travels 5 km north, then 3 km south. Find:

(a) the total distance travelled

(b) the displacement (taking north as positive)

(a) Total distance = 5 + 3 = 8 km

(b) Displacement = +5 + (−3) = +2 km (or 2 km north)

Practice

Question 2: A particle has velocity −4 m/s and acceleration +2 m/s². Describe its motion in words.

The particle is moving in the negative direction (velocity is negative).

The acceleration is positive, meaning velocity is increasing (becoming less negative).

Since velocity and acceleration have opposite signs, the particle is slowing down.

AnswerThe particle is moving in the negative direction and decelerating (slowing down).

Practice

Question 3: Express the velocity of an object moving at 15 m/s at 30° above the horizontal using i-j notation.

Horizontal (i) component:

15 cos 30° = 15 \times \frac{3}{2} = \frac{15 3}{2} \approx 13.0

m/s

Vertical (j) component:

15 sin 30° = 15 \times 0.5 = 7.5

m/s

Answer: $v = (13.0 i + 7.5 j)$ m/s

Or in exact form:

v = (\frac{15 3}{2} i + 7.5 j)

m/s

Summary

Key Concepts

ScalarVector

Distance (total path length)Displacement (change in position)

Speed (rate of distance change)Velocity (rate of displacement change)

Always positiveAcceleration (rate of velocity change)

Key Exam Skills

Always define your positive direction

Convert units carefully (km/h ↔ m/s: divide or multiply by 3.6)

Remember: opposite signs of v and a means slowing down

Use i-j notation for 2D problems