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Acceleration as a function of time

Edexcel IAL Mathematics · WME03 Mechanics 3

In M1 the acceleration was constant, so suvat worked. In M3 the acceleration changes with time, so you use calculus instead: differentiate to go from displacement to velocity to acceleration, and integrate to go back.

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The calculus chain

For motion in a straight line:

Every integration gives a constant. Find it from the initial conditions in the question, for example "P starts from rest at O" means v = 0 and x = 0 when t = 0.

Functions you must handle

Pearson says the calculus will match P1 to P4, so a(t) or v(t) may be a polynomial, an exponential such as 8e-0.5t, a trig function such as sin(πt/4) (in radians) or a fraction such as 12/(t + 1)2. Integrate exactly, then substitute.

Distance or displacement?

Displacement is ∫v dt between the limits. Distance travelled is ∫|v| dt: if v changes sign, find the time it is zero, integrate each part separately and add the sizes. A limiting (terminal) speed is the value v approaches as t gets very large, for example when e-t tends to 0.

Worked example

A particle P moves on a straight line. At time t seconds its acceleration is (4 - 2t) m/s2. When t = 0, P is at O with velocity 3 m/s. Find v and x in terms of t, and the time when P first comes to rest.

  1. v = ∫(4 - 2t) dt = 4t - t2 + c; v = 3 at t = 0, so v = 3 + 4t - t2.
  2. x = ∫v dt = 3t + 2t2 - t3/3 (+0, since x = 0 at t = 0).
  3. Rest: t2 - 4t - 3 = 0, so t = 2 + sqrt 7 = 4.65 (the negative root is rejected).

Answer: v = 3 + 4t - t2, x = 3t + 2t2 - t3/3, rest at t = 4.65 s

Key idea: Integrate a to get v and v to get x, using the starting values to find each constant. For distance, split the interval where v = 0.

Check yourself

  1. A particle has acceleration (6t - 2) m/s2 and velocity 1 m/s when t = 0. Find its velocity when t = 3.
    Show answer

    22 m/s. v = 3t2 - 2t + 1, so v(3) = 27 - 6 + 1 = 22 m/s.

  2. A particle moves with velocity v = 8e-0.5t m/s. Find its acceleration when t = 2, to 3 significant figures.
    Show answer

    -1.47 m/s2. a = dv/dt = -4e-0.5t, so a(2) = -4e-1 = -1.47 m/s2 (it is slowing down).

  3. For the same particle (v = 8e-0.5t m/s), find the distance travelled in the first 4 seconds, to 3 significant figures.
    Show answer

    13.8 m. v > 0 throughout, so distance = ∫ from 0 to 4 of 8e-0.5t dt = 16(1 - e-2) = 13.8 m.

  4. A particle has velocity v = 2 sin(πt/4) m/s. Find its displacement from its starting point after 2 seconds, to 3 significant figures.
    Show answer

    2.55 m. x = ∫ from 0 to 2 of 2 sin(πt/4) dt = (8/π)[-cos(πt/4)] from 0 to 2 = (8/π)(1 - 0) = 2.55 m.

  5. A particle starts from rest. Its acceleration is 12/(t + 1)2 m/s2. Its speed approaches a limiting value as t increases. Find it.
    Show answer

    12 m/s. v = 12 - 12/(t + 1) (using v = 0 at t = 0). As t gets large, 12/(t + 1) tends to 0, so v tends to 12 m/s.

  6. A particle has velocity v = (t2 - 5t + 4) m/s. Find the total distance it travels between t = 0 and t = 4, to 3 significant figures.
    Show answer

    6.33 m. v = (t - 1)(t - 4) changes sign at t = 1. From 0 to 1 it moves 1/3 - 5/2 + 4 = 11/6 m forwards; from 1 to 4 the integral is -9/2, so 9/2 m back. Distance = 11/6 + 9/2 = 6.33 m.

  7. The acceleration of a particle is given as a function of t. How do you find its velocity?
    • Differentiate a with respect to t
    • Use v = u + at with the value of a at the start
    • Integrate a with respect to t and use an initial condition
    • Multiply a by t
    Show answer

    Integrate a with respect to t and use an initial condition. v = ∫a dt; the constant comes from a known value of v.

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