Overview

Linear motion is motion along a straight line. We study how objects move, how fast they move, and how their speed changes. We use simple measurements (distance, time) and equations that apply when acceleration is constant. In Kenya, examples include buses slowing at matatu stops, a ball dropped from a roof, or a bicycle starting from rest.

Specific Learning Outcomes

• a) Explain terms used in linear motion (distance, displacement, speed, velocity, acceleration, scalar and vector).
• b) Calculate variables of motion using equations of linear motion with constant acceleration (use v = u + at, s = ut + ½at², v² = u² + 2as).
• c) Investigate the effect of gravity on bodies under free fall (use g ≈ 9.8 m/s²; for school problems g = 10 m/s² may be used).
• d) Appreciate the significance of linear motion and free fall in real life (transport, sports, safety, engineering).

Key terms (simple definitions)

Distance

How much ground an object has covered (scalar, unit: metre, m).

Displacement

Change in position from start to finish (vector, unit: m). Example: start at 0 m, finish at +5 m → displacement +5 m.

Speed

Distance travelled per unit time (scalar). Average speed = total distance / time (m/s).

Velocity

Displacement per unit time (vector). Average velocity = displacement / time (m/s).

Acceleration

Rate of change of velocity (vector, m/s²). If speed increases forward → positive acceleration.

Free fall

Motion under gravity alone (no air resistance). Acceleration = g (≈ 9.8 m/s² downward).

Important equations (constant acceleration)

Use consistent sign convention (choose up or right as positive). Common equations:

• v = u + a t
• s = u t + ½ a t²
• v² = u² + 2 a s

Where: u = initial velocity (m/s), v = final velocity (m/s), a = acceleration (m/s²), t = time (s), s = displacement (m).

Worked examples (step-by-step)

Simple visual expressions

Suggested learning experiences (hands-on & classroom)

1. Simple drop test (free fall):
   • Materials: small ball, meter rule, stopwatch or smartphone slow motion camera.
   • Method: Measure height, drop ball (release without push), time fall. Repeat several times and calculate average time. Use s = ½ g t² to estimate g from data.
   • Notes: Use small heights (1–3 m) and take care; do outside or in corridor. Explain experimental error (reaction time, air resistance).
2. Inclined plane for constant acceleration:
   • Roll a toy car down a wooden plank set at different angles to show smaller acceleration than free fall.
   • Measure distance and time or mark equal time positions and observe increasing spacing between marks.
3. Graphing activity:
   • Collect data from an accelerating object (e.g., stopwatch and measured distances) and plot displacement–time and velocity–time graphs on graph paper or using a computer.
4. Problem solving:
   • Work in pairs to solve motion problems (using v = u + at, s = ut + ½at²). Swap solutions and explain steps to the class to build reasoning skills.

Safety: When doing drops use safety goggles if objects break; do not stand under falling objects; choose soft landing areas for experiments.

Real-life applications (Kenyan context)

• Road safety: understanding braking distance when a matatu or private car decelerates helps set safe following distance.
• Sports: coaches use motion ideas in sprint starts, long jump, shot put and ball trajectories.
• Construction: predicting how fast an object falls for safe site planning and scaffold heights.
• Everyday: using elevator motion, bicycle acceleration and deceleration, and estimating time to stop or reach a destination.

Assessment questions (practice)

1. A bicycle accelerates from 2 m/s to 8 m/s in 3 s. Find the acceleration and the distance covered in this time.
2. A stone thrown upwards at 20 m/s. Using g = 10 m/s², find time to return to hand and maximum height.
3. Drop experiment data: height = 1.8 m, measured average fall time = 0.6 s. Calculate g from the data and comment on difference from 9.8 m/s².

Answers (brief)

1. a = (8-2)/3 = 2 m/s². s = ut + ½at² = 2×3 + 0.5×2×9 = 6 + 9 = 15 m.
2. Time to top t = u/g = 20/10 = 2 s, time to return = 4 s total. Height = u²/(2g) = 400/(20) = 20 m.
3. Use s = ½ g t² → g = 2s/t² = (2×1.8)/(0.6²) = 3.6/0.36 = 10 m/s². The result is higher than 9.8 because of rounding and measurement error (stopwatch reaction time, air resistance small at low heights).

Tips for learners

• Always state sign convention before solving (which direction is positive).
• Check units: convert cm to m, minutes to seconds before substituting in formulas.
• Use g = 9.8 m/s² for accurate results; many class problems use g = 10 m/s² for simpler arithmetic.
• When throwing objects up, acceleration = -g if up is positive.