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Distance travelled when a particle changes direction

Edexcel IAL Mathematics · WME02 Mechanics 2

Distance and displacement are different when a particle turns round. This is one of the most common places to lose marks in M2, and it also covers motion described by two different formulae over two time intervals.

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Displacement or distance?

Displacement is the change in position, ∫ v dt over the interval. Distance is the total length travelled, always positive. If v changes sign in the interval, the particle turned round, and the integral alone gives the wrong distance.

Method for distance

1. Solve v = 0 to find when the particle turns round (not x = 0: Pearson's examiners report this mistake).

2. Split the interval at each turning time.

3. Find the displacement over each part and add their sizes.

A quick sketch of v against t shows which parts are below the axis.

Two-part motion

Some questions give one formula for 0 ≤ t ≤ T and another for t > T. Velocity is usually continuous, so the two formulae give the same v at t = T: this often finds an unknown constant. Work out the distance for each part separately and add.

Worked example

A particle moves with v = 3t2 - 12t + 9 m/s, starting at O. Find the distance it travels in the first 4 seconds.

  1. v = 3(t - 1)(t - 3), so it turns round at t = 1 and t = 3.
  2. x = t3 - 6t2 + 9t: x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4.
  3. Distances: 4 (out), then 4 (back to O), then 4 (out again).

Answer: Distance = 12 m (the displacement is only 4 m)

Key idea: Solve v = 0, split the interval there, and add the sizes of the displacements in each part.

Check yourself

  1. A particle moves with v = 6 - 2t m/s. Find the distance it travels in the first 5 seconds.
    Show answer

    13 m. v = 0 at t = 3. x = 6t - t2: x(3) = 9, x(5) = 5. Distance = 9 + (9 - 5) = 13 m.

  2. For the same particle (v = 6 - 2t), find its displacement from its start after 5 seconds.
    Show answer

    5 m. ∫ from 0 to 5 of (6 - 2t) dt = 30 - 25 = 5 m.

  3. A particle moves with v = t2 - 6t + 8 m/s. Find the distance it travels between t = 0 and t = 4.
    Show answer

    8 m. v = (t - 2)(t - 4), so it turns at t = 2. x = t3/3 - 3t2 + 8t: x(2) = 20/3, x(4) = 16/3. Distance = 20/3 + 4/3 = 8 m.

  4. A particle moves with v = 3t2 for 0 ≤ t ≤ 2 and v = 16 - 2t for t > 2 (m/s). Find the distance travelled from t = 0 to t = 8.
    Show answer

    44 m. First part: ∫ 3t2 from 0 to 2 = 8. Second part: ∫ (16 - 2t) from 2 to 8 = (128 - 64) - (32 - 4) = 36, and v ≥ 0 throughout. Total 44 m.

  5. A particle moves with x = 6t2 - t3 metres. Find the time t > 0 when it returns to its starting point.
    Show answer

    6 s. x = t2(6 - t) = 0 gives t = 6.

  6. A particle has v = kt2 for 0 ≤ t ≤ 2 and v = 20 - 4t for t > 2 (m/s). Velocity is continuous at t = 2. Find k.
    Show answer

    3. At t = 2: 4k = 20 - 8 = 12, so k = 3.

  7. A question asks for the total distance travelled in the first 4 s. What should you do first?
    • Find when x = 0
    • Find any times in 0 < t < 4 when v = 0
    • Integrate v from 0 to 4 and stop there
    • Find when a = 0
    Show answer

    Find any times in 0 < t < 4 when v = 0. Distance needs the turning points, where v = 0.

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Last updated 8 November 2026. Spotted a mistake? Tell us.