Calculate total distance when a particle changes direction
Find every velocity zero in the stated interval, test for reversal and add the magnitudes of displacement on each motion segment.
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
- Solve linear and quadratic velocity equations.
- Evaluate a polynomial primitive at interval endpoints.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For v = 4 − 2t on 0 ≤ t ≤ 3, where should the interval be split for distance?
What you will learn
- Locate and verify all reversals in a given time interval.
- Calculate distance and displacement separately with graph areas or integration.
Split the interval where velocity changes sign
Displacement is the signed integral of velocity, but total distance is the integral of |v|. A practical Paper 1 method is to solve v = 0, retain roots within the time interval, inspect signs, and evaluate displacement separately on intervals of one direction. Add the absolute values of those segment displacements. A velocity zero without a sign change is momentary rest and does not require a new direction, though splitting there is harmless.
For v = 6 − 2t on 0 ≤ t ≤ 5, velocity changes from positive to negative at 3 s. A primitive is S(t) = 6t − t². Its values are S(0) = 0, S(3) = 9 and S(5) = 5 m. Thus the segment displacements are +9 and −4 m. Their signed sum gives displacement 5 m, while their magnitudes give distance 13 m. Taking |S(5) − S(0)| would miss the four metres of negative travel.
Multiple reversals and a non-zero origin need careful bookkeeping
For v = t² − 4t + 3 = (t − 1)(t − 3), the particle reverses at both 1 and 3 s on [0, 4]. A primitive is S = t³/3 − 2t² + 3t. It has values 0, 4/3, 0 and 4/3 at 0, 1, 3 and 4 s. The three displacements are +4/3, −4/3 and +4/3 m; total distance is 4 m although displacement is only 4/3 m.
The additive initial position cancels from every segment change, so a non-zero starting coordinate does not alter this method. Once distance and displacement are known, divide each by the whole elapsed time for average speed and average velocity respectively. Include units and avoid rounding intermediate turning positions: exact roots and fractions preserve accuracy before the requested final rounding. Check that total distance is at least the magnitude of displacement.
Pause and explain
Segment displacements are +4/3, −4/3 and +4/3 m. What is total distance?
Worked example
For v = 6 − 2t m s⁻¹ on 0 ≤ t ≤ 5 s, calculate distance, displacement and both whole-journey averages.
Show the worked solution
- Solve v = 0 to get reversal at 3 s; the primitive S = 6t − t² has values 0, 9 and 5 m.
- The particle travels 9 m positively then 4 m negatively, giving distance 13 m and displacement +5 m.
- Divide by 5 s: average speed is 2.6 m s⁻¹ and average velocity is +1 m s⁻¹.
Answer Distance 13 m; displacement +5 m; average speed 2.6 m s⁻¹; average velocity +1 m s⁻¹.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
For v = 4 − 2t m s⁻¹ over 0 ≤ t ≤ 3 s, find displacement and total distance.
Give me a hint
Evaluate the primitive at the start, reversal and end.
Compare my reasoning
- The reversal is t = 2 s and a primitive is S = 4t − t².
- S(0) = 0, S(2) = 4 and S(3) = 3 m, giving segment changes +4 and −1 m.
- Displacement is +3 m and distance is 4 + 1 = 5 m.
Displacement +3 m; distance 5 m.
Look for these in your work
- Verify the velocity sign changes at 2 s.
- Add segment magnitudes for the distance.
For v = t² − 4t + 3 m s⁻¹ on 0 ≤ t ≤ 4 s, find reversals, distance, displacement, average speed and average velocity.
Give me a hint
There are two roots inside the interval, creating three one-direction stages.
Compare my reasoning
- Factor to get reversal times 1 and 3 s; use S = t³/3 − 2t² + 3t.
- At 0, 1, 3 and 4 s, S is 0, 4/3, 0 and 4/3 m, so distance is 4 m and displacement is 4/3 m.
- Over 4 s, average speed is 1 m s⁻¹ and average velocity is +1/3 m s⁻¹.
Reversals at 1 and 3 s; distance 4 m; displacement +4/3 m; average speed 1 m s⁻¹; average velocity +1/3 m s⁻¹.
Look for these in your work
- Evaluate all four interval endpoints exactly.
- Check distance exceeds the displacement magnitude and use total elapsed time.
A particle’s position is s = t² − 6t + 10 m for 0 ≤ t ≤ 5 s. Find its reversal, distance, displacement and both average rates.
Give me a hint
The 10 m initial coordinate cancels from changes but must be included in endpoint positions.
Compare my reasoning
- Velocity is 2t − 6, so reversal occurs at 3 s.
- Positions at 0, 3 and 5 s are 10, 1 and 5 m; distance is 9 + 4 = 13 m and displacement is 5 − 10 = −5 m.
- Average speed is 13/5 = 2.6 m s⁻¹ and average velocity is −5/5 = −1 m s⁻¹.
Reversal 3 s; distance 13 m; displacement −5 m; average speed 2.6 m s⁻¹; average velocity −1 m s⁻¹.
Look for these in your work
- Use the initial coordinate rather than assuming s(0) = 0.
- Separate the path length from the negative net change.
Common mistakes
- Absolute net displacement is not total distance after a reversal.
- Keep only turning times within the stated interval.
- Use all motion segments and postpone rounding their endpoints.
What can you explain now?
Why is |∫₀⁵ v dt| insufficient for distance when v changes sign?
Compare with the explanation
Positive and negative travel cancel in the integral before its magnitude is taken.
Distance must add the magnitude of each one-direction segment instead.
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Make it stick
Solve v = 0, inspect signs, evaluate every segment displacement and add magnitudes for distance.
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.