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Variable acceleration: integrating to find velocity and displacement

Edexcel IAL Mathematics · WME02 Mechanics 2

Going the other way, you integrate acceleration to get velocity and integrate velocity to get displacement. Each integration brings a constant, which you find from information given in the question.

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Integrating

Each indefinite integral needs + c. Use the starting conditions, such as "starts from rest" (v = 0 when t = 0) or "starts at O" (x = 0 when t = 0), to find c. Pearson's examiners note that marks are lost when candidates do not show why c = 0: write the substitution out.

Definite integrals

The displacement between t = t_1 and t = t_2 is ∫ v dt from t_1 to t_2. This gives the change in position, not the distance travelled if the particle turns round (see the next lesson).

Using both directions

A question may give a, ask for v, then ask when the particle is at rest or where it is then. Keep your v and x functions written out in full and label them, since later parts depend on them.

Worked example

A particle starts from O with velocity 3 m/s and moves with acceleration a = 6t - 4 m/s2. Find its velocity and displacement when t = 2.

  1. v = 3t2 - 4t + c. At t = 0, v = 3, so c = 3: v = 3t2 - 4t + 3.
  2. x = t3 - 2t2 + 3t + k. At t = 0, x = 0, so k = 0.
  3. At t = 2: v = 12 - 8 + 3 = 7 m/s, x = 8 - 8 + 6 = 6 m.

Answer: v = 7 m/s, x = 6 m

Key idea: Integrate a for v, integrate v for x, and find each constant from the given conditions. A definite integral of v gives displacement.

Check yourself

  1. A particle starts from rest and moves with acceleration a = 4t m/s2. Find its velocity when t = 3.
    Show answer

    18 m/s. v = 2t2 + c and v = 0 at t = 0, so c = 0. v(3) = 18 m/s.

  2. A particle starts at O and moves with v = 3t2 + 2 m/s. Find its displacement from O when t = 2.
    Show answer

    12 m. x = t3 + 2t + c with c = 0, so x(2) = 8 + 4 = 12 m.

  3. A particle starts from rest and moves with a = 6 - 2t m/s2. Find its greatest velocity.
    Show answer

    9 m/s. v = 6t - t2 (c = 0). Greatest when a = 0, t = 3: v = 18 - 9 = 9 m/s.

  4. A particle moves with v = 6t2 - 2t m/s. Find its displacement between t = 1 and t = 3.
    Show answer

    44 m. ∫ (6t2 - 2t) dt = [2t3 - t2] from 1 to 3 = (54 - 9) - (2 - 1) = 44 m.

  5. A particle starts from rest and moves with a = 2e0.5t m/s2. Find its velocity when t = 2, to 3 significant figures.
    Show answer

    6.87 m/s. v = 4e0.5t + c; v = 0 at t = 0 gives c = -4. v(2) = 4e - 4 = 6.87 m/s.

  6. A particle moves with a = 2t - 5 m/s2 and has velocity 6 m/s when t = 0. Find the times when it is at rest.
    Show answer

    t = 2 s and t = 3 s. v = t2 - 5t + 6 = (t - 2)(t - 3), so v = 0 at t = 2 and t = 3.

  7. A particle's velocity is found by integrating its acceleration. How is the constant of integration found?
    • It is always zero
    • From the value of a at t = 0
    • From a known value of v at a particular time
    • By differentiating again
    Show answer

    From a known value of v at a particular time. You need a condition on v, such as v = 3 when t = 0.

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