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Core Mathematics — Measurements & Geometry

Subtopic: Linear Motion (Age 15, Kenyan context)

Specific learning outcomes

(a) Explain distance, displacement, speed, velocity and acceleration with real-life examples.
(b) Determine velocity and acceleration in different situations (calculations).
(c) Draw displacement–time (s–t) graphs for situations.
(d) Interpret displacement–time graphs.
(e) Draw velocity–time (v–t) graphs from tables.
(f) Interpret velocity–time graphs.
(g) Determine relative speed of two moving bodies.
(h) Appreciate uses of linear motion in everyday life (transport, sports, engineering).

Key terms (with simple Kenyan examples)

Distance: How much ground is covered, regardless of direction. Example: walking 1.2 km from home to a shop and then 0.8 km back → total distance = 2.0 km.
Displacement: Change in position from start to finish, with direction. Example: If your home to school is 1 km east and you stop at school, displacement = +1 km (east). If you return home, displacement = 0.
Speed: How fast an object moves (scalar). Speed = distance ÷ time. Units: m/s or km/h. Example: a boda boda travelling 36 km/h.
Velocity: Speed with direction (vector). Velocity = displacement ÷ time. Example: 3 m/s east.
Acceleration: Rate of change of velocity. Acceleration = (change in velocity) ÷ time. Units: m/s². Example: a matatu accelerating from stop to 20 m/s in 10 s → acceleration = 2 m/s².

Important formulas and unit conversions

Speed = distance / time. v = s / t
Velocity = displacement / time. v = Δx / t
Acceleration = change in velocity / time. a = Δv / t
Unit conversion: 1 km/h = 5/18 m/s. To convert km/h → m/s multiply by 5/18.
Area under v–t graph = displacement (signed).
Slope of s–t graph = velocity.

Worked examples

Example 1 — Constant speed
A student walks from home to school, distance 1200 m, in 20 minutes. Find the average speed in m/s.
Solution: 20 minutes = 1200 s. Speed = distance / time = 1200 m / 1200 s = 1.0 m/s.

Example 2 — Velocity and displacement
A runner sprints 100 m east in 12 s, then 50 m west in 6 s. Find total distance, displacement and average velocity for the whole run.
Distance = 100 + 50 = 150 m.
Displacement = 100 m east − 50 m west = 50 m east.
Total time = 12 + 6 = 18 s. Average velocity = displacement / time = 50 / 18 ≈ 2.78 m/s east.

Example 3 — Acceleration
A matatu accelerates from rest to 20 m/s in 10 s. Determine acceleration.
a = Δv / t = (20 − 0) / 10 = 2.0 m/s².

Drawing and interpreting displacement–time (s–t) graphs

Axes: horizontal = time (s or min), vertical = displacement (m or km). Slope = velocity (positive slope → moving away; zero slope → stationary; negative slope → returning).

Constant speed — straight line

Horizontal = stopped; sloped = moving

Positive then negative slope = return

How to interpret: calculate slope (change in s / change in t). Slope positive → moving away; slope zero → stopped; slope negative → coming back. Units of slope = m/s (or km/min).

Velocity–time (v–t) graphs: drawing and interpretation

Axes: horizontal = time, vertical = velocity. Slope = acceleration. Area under v–t = displacement.

Flat line = constant velocity

Sloped line = changing velocity (acceleration)

Below axis = negative velocity (opposite direction)

Example: A car has velocity 10 m/s for 5 s, then 0 m/s for 2 s. Displacement = area of rectangle = 10 × 5 = 50 m.

Relative speed (two bodies)

When two bodies move:

Same direction: relative speed = |v1 − v2| (they may overtake).
Opposite directions: relative speed = v1 + v2 (they approach each other faster).

Example — Opposite direction (meeting):
A boda boda travels 20 m/s north and a matatu travels 15 m/s south. How fast are they approaching each other? 20 + 15 = 35 m/s.

Example — Same direction (overtaking):
A car at 90 km/h and a lorry at 60 km/h on the Nairobi–Mombasa road. Convert to m/s: 90 × 5/18 = 25 m/s, 60 × 5/18 = 16.67 m/s. Relative speed = 25 − 16.67 = 8.33 m/s. Time to overtake if car is initially 50 m behind: t = distance / relative speed = 50 / 8.33 ≈ 6 s.

Classroom & field activities (Suggested learning experiences)

Measure walking speed: In groups, mark 50 m on the school ground. One student walks at normal pace; another times with a stopwatch. Compute speed (m/s) and convert to km/h.
Plot s–t graph: Use the times and displacements measured and draw a displacement–time graph on graph paper. Identify when the student stopped (horizontal) or returned (negative slope if direction reversed).
Vehicle observation (real-life): Use a smartphone stopwatch to time a boda boda or matatu over a known distance. Calculate average speed and discuss safety implications of high acceleration/deceleration.
Table → v–t graph: Give students a table of velocities at different times (e.g., 0s:0, 2s:4, 4s:8, 6s:8, 8s:0). Ask them to draw the v–t graph and calculate displacement by finding areas.
Relative speed problem: Role-play two students walking towards each other from opposite ends; measure time to meet and compare with calculation using relative speed formulas.

Practice exercises (with short answers)

A student cycles 3 km to school in 12 minutes. Find speed in km/h and m/s. (Answer: 15 km/h; 15 × 5/18 = 4.17 m/s)
A vehicle moves 200 m east then 150 m west. If this takes 70 s, find distance, displacement and average velocity. (Answer: distance = 350 m; displacement = 50 m east; avg velocity = 50/70 ≈ 0.71 m/s east)
From a table: at t = 0, v = 0; t = 5 s, v = 10 m/s. Find acceleration and displacement in the first 5 s. (Answer: a = 2 m/s²; displacement = area of triangle = 0.5 × 5 × 10 = 25 m)
Two cars start 180 km apart and drive towards each other at 60 km/h and 40 km/h. How long until they meet? (Answer: relative speed = 100 km/h → time = 180/100 = 1.8 h = 1 h 48 min)

Why learn linear motion? (Appreciation)

Transport planning: speeds and travel times for buses, matatus, trains.
Safety: understanding braking distances (deceleration) helps design safer roads and make sound decisions when driving or cycling.
Sports: coaches use velocity and acceleration to improve sprint starts and pacing.
Engineering and design: machines and structures require knowledge of motion (e.g., conveyor belts, elevators).

Quick summary

Distance is how much ground covered. Displacement is change in position. Speed = distance/time. Velocity = displacement/time (direction matters). Acceleration = change in velocity/time. Use s–t graphs to find velocity (slope) and v–t graphs to find acceleration (slope) and displacement (area).

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