Original revision notes and practice for selected chapter concepts. Chapter sequence: PECTAA-listed English-medium Physics 9, revised NCP 2023. Not an official exercise key or an exam-session smart syllabus.

Distance and displacement

Distance measures the length of the travelled path and is a scalar. Displacement measures the change from the starting position to the final position and includes direction. An object can travel a non-zero distance and return to zero displacement. State the reference frame when describing whether something is moving.

  • For motion on a line, choose a positive direction before using signed positions.
  • A return to the starting position gives zero net displacement.
  • Translational, rotational and vibratory descriptions refer to different patterns of motion.

Speed, velocity and acceleration

Average speed divides total distance by elapsed time. Average velocity divides net displacement by the same elapsed time. Instantaneous values describe motion at a particular moment. Acceleration measures change in velocity per unit time, so a change in speed or direction can involve acceleration.

  • Average speed = total distance ÷ elapsed time.
  • Average velocity = displacement ÷ elapsed time.
  • Average acceleration = (final velocity − initial velocity) ÷ elapsed time; its SI unit is m/s².

Use the motion equations only with their conditions

For straight-line motion with constant acceleration, use u for initial velocity, v for final velocity, a for acceleration, t for time and s for displacement. Choose the equation that contains the known values and the required unknown. Signed quantities must follow the same direction convention throughout.

  • v = u + at
  • s = ut + ½at²
  • v² = u² + 2as
  • For free fall near Earth, neglecting air resistance, use g ≈ 9.8 m/s² downward unless another value is specified.

Read motion graphs from their axes

The gradient of a distance–time graph describes speed. The gradient of a velocity–time graph describes acceleration, and its signed area describes displacement. A speed–time area gives distance. Check the axis labels before deciding what a gradient or area means.

  • A horizontal velocity–time line means constant velocity, not rest unless its value is zero.
  • A negative velocity is motion in the chosen negative direction.
  • Negative acceleration does not always mean slowing down; compare the signs of velocity and acceleration.
Velocity–time graph for constant accelerationVelocity rises linearly from zero to 10 metres per second in 5 seconds. The slope is 2 metres per second squared and the triangular area is 25 metres. Velocity (m/s)Time (s)0105Area = 25 m
Original example: constant acceleration from 0 to 10 m/s in 5 s. Gradient = 10 ÷ 5 = 2 m/s²; displacement = ½ × 5 × 10 = 25 m.

Worked examples

These are original SchoolOrbit examples. They are not numbered textbook exercises.

One journey, two average rates

A learner walks 60 m east, then 20 m west, taking 20 s in total. Find the distance, displacement, average speed and average velocity.

  1. Distance = 60 + 20 = 80 m.
  2. With east positive, displacement = 60 − 20 = 40 m east.
  3. Average speed = 80 ÷ 20 = 4 m/s.
  4. Average velocity = 40 ÷ 20 = 2 m/s east.

Result: 80 m; 40 m east; 4 m/s; 2 m/s east

Constant acceleration in a straight line

A trolley starts at 2 m/s and accelerates at 3 m/s² for 4 s in the same direction. Find its final velocity and displacement.

  1. v = u + at = 2 + 3 × 4 = 14 m/s.
  2. s = ut + ½at² = 2 × 4 + ½ × 3 × 4².
  3. s = 8 + 24 = 32 m.

Result: 14 m/s and 32 m

Try the original practice

Work out each answer first, then open the explanation. This page does not record answers or award an online score.

Practice 1

A runner completes one lap and returns to the starting point. Which net displacement is correct?

  1. The lap length
  2. Half the lap length
  3. Zero
  4. The elapsed time
Show answer and explanation

Zero

The final and initial positions are identical, although the travelled distance is not zero.

Practice 2

A car changes velocity from 4 m/s to 16 m/s in 6 s with constant acceleration. Find its acceleration.

Show answer and explanation

2 m/s²

a = (16 − 4) ÷ 6 = 2 m/s².

Practice 3

A vehicle moves at a constant speed of 6 m/s for 5 s. Find its distance.

Show answer and explanation

30 m

Distance = speed × time = 6 × 5 = 30 m. This is also the area under its speed–time graph.

Common mistakes to avoid

  • Using distance in an average-velocity calculation.
  • Applying constant-acceleration equations when acceleration varies.
  • Calling every negative acceleration a decrease in speed.

Sources and scope

The chapter sequence is checked against the official directory and source book. The explanations, numerical values and practice are independently written for SchoolOrbit. No PECTAA endorsement, complete exercise key or examination-session coverage is claimed.

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