Distinguish distance displacement speed and directed velocity
Use a signed line to describe motion, calculate two different journey averages and decide when acceleration increases or decreases speed.
Section 4.2 · Kinematics of motion in a straight line
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Add signed quantities on a number line.
- Use metres, seconds and rates with consistent units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Taking right as positive, what displacement results from 5 m right then 8 m left?
What you will learn
- Calculate distance, displacement, average speed and average velocity.
- Interpret the signs of velocity and acceleration without confusing direction with speed.
A journey has both a length and a directed change
Choose an origin and positive direction along the line. Position x locates the particle relative to that origin; displacement over a journey is final position minus initial position. Distance adds the lengths of all parts travelled and cannot be negative. Travelling 6 m right and then 2 m left gives displacement +4 m but distance 8 m. Returning to the starting point gives zero displacement even after a long journey.
Average velocity is displacement divided by elapsed time, while average speed is distance divided by elapsed time. If that 8 m journey takes 4 s, average speed is 2 m s⁻¹ and average velocity is +1 m s⁻¹. Neither is necessarily the instantaneous value at a particular moment. Do not average two segment speeds without considering the time spent at each speed; the definition uses the whole journey’s distance and time.
A negative velocity can describe increasing or decreasing speed
Velocity is the directed rate of change of position, and speed is its magnitude |v|. Acceleration is the directed rate of change of velocity. At a moment when v is non-zero, matching signs of v and a increase speed, while opposite signs decrease speed. A particle with v = −8 m s⁻¹ and a = +2 m s⁻² slows initially: its negative velocity approaches zero rather than becoming more negative.
With that constant acceleration, v = −8 + 2t and rest occurs at t = 4 s. The particle then moves in the positive direction and speeds up if the acceleration continues. Zero velocity at an instant does not imply zero acceleration or permanent rest. The term deceleration concerns decreasing speed; an acceleration can be positive while the particle decelerates, and negative while it speeds up. Restrict these descriptions to the specified time interval.
Pause and explain
If v is negative and a is positive, what happens to speed before v reaches zero?
Worked example
A particle travels 6 m right then 2 m left in 4 s. Taking right as positive, calculate distance, displacement and both averages.
Show the worked solution
- Distance is 6 + 2 = 8 m; displacement is 6 − 2 = +4 m.
- Average speed is 8/4 = 2 m s⁻¹.
- Average velocity is 4/4 = +1 m s⁻¹, directed rightwards.
Answer Distance 8 m; displacement +4 m; average speed 2 m s⁻¹; average velocity +1 m s⁻¹.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle travels 11 m right then 5 m left in 8 s. Take right as positive. Find distance, displacement, average speed and average velocity.
Give me a hint
Add path lengths for distance and use signs for displacement.
Compare my reasoning
- Distance is 11 + 5 = 16 m, while displacement is 11 − 5 = +6 m.
- Average speed is 16/8 = 2 m s⁻¹.
- Average velocity is 6/8 = +0.75 m s⁻¹.
Distance 16 m; displacement +6 m; average speed 2 m s⁻¹; average velocity +0.75 m s⁻¹.
Look for these in your work
- Keep the journey length distinct from the directed change.
- Use the same 8 s elapsed time and state the velocity sign.
A particle moves from coordinate 4 m to −2 m, then to 1 m, in a total of 6 s. Find the two journey averages.
Give me a hint
Compute the length of each segment before dividing by total time.
Compare my reasoning
- The segment changes are −6 m and +3 m, giving distance 6 + 3 = 9 m.
- Total displacement is 1 − 4 = −3 m.
- Average speed is 9/6 = 1.5 m s⁻¹ and average velocity is −3/6 = −0.5 m s⁻¹.
Average speed 1.5 m s⁻¹; average velocity −0.5 m s⁻¹.
Look for these in your work
- Use initial and final coordinates for displacement.
- Add segment lengths even though their signed changes partly cancel.
For 0 ≤ t ≤ 6 s, a particle has v = −8 + 2t m s⁻¹. State its direction and whether speed increases or decreases before and after rest; find speeds at t = 0, 4 and 6.
Give me a hint
Solve v = 0 first, then compare the signs of velocity and acceleration.
Compare my reasoning
- Acceleration is +2 m s⁻² and rest occurs at t = 4 s.
- For 0 ≤ t < 4, v < 0 and speed decreases; for 4 < t ≤ 6, v > 0 and speed increases.
- Speeds are |−8| = 8, 0 and |4| = 4 m s⁻¹ at the stated times.
Negative direction and decreasing speed before 4 s; positive direction and increasing speed after 4 s; speeds 8, 0 and 4 m s⁻¹.
Look for these in your work
- Describe the instant of rest separately from its neighbouring intervals.
- Use |v| for speed rather than the signed velocity.
Common mistakes
- Distance is not the magnitude of displacement when the direction changes.
- Use elapsed time for both journey averages.
- Negative acceleration does not always mean decreasing speed.
What can you explain now?
A particle has v = −3 m s⁻¹ and a = −2 m s⁻². Is its speed increasing at that instant?
Compare with the explanation
Yes, its speed is increasing.
The negative velocity becomes more negative, so its magnitude increases.
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Make it stick
Choose a signed direction, separate distance from displacement and compare both velocity and acceleration signs when describing speed.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.