Variable acceleration: differentiating displacement

The chain

If x is the displacement from a fixed point O at time t:

• v = dx/dt
• a = dv/dt = d^2x/dt².

The functions may be polynomials, exponentials or trigonometric functions: any calculus from P1 to P4 can appear.

Key conditions

• At rest (instantaneously): v = 0.
• At O: x = 0.
• Maximum or minimum velocity: dv/dt = 0, so a = 0.
• The particle changes direction when v = 0 and v changes sign.

Always check that the times you find are in the range the question allows (for example t ≥ 0).

Do not use suvat

Pearson's examiners report that some candidates use suvat in variable acceleration questions. If a, v or x is given as a function of t, the acceleration is not constant, so use calculus only.

Worked example

A particle moves along a straight line so that x = t³ - 6t² + 9t metres at time t seconds. Find when it is at rest and its acceleration at those times.

1. v = dx/dt = 3t² - 12t + 9 = 3(t - 1)(t - 3).
2. At rest when t = 1 and t = 3.
3. a = dv/dt = 6t - 12, so a = -6 m/s² at t = 1 and a = 6 m/s² at t = 3.

Answer: At rest at t = 1 s (a = -6 m/s²) and t = 3 s (a = 6 m/s²)

Check yourself

1. A particle moves on a line with x = 2t³ - 3t² metres. Find its velocity when t = 2.
   Show answer

   12 m/s. v = 6t² - 6t, so v(2) = 24 - 12 = 12 m/s.

2. For the same particle (x = 2t³ - 3t²), find its acceleration when t = 2.
   Show answer

   18 m/s². a = 12t - 6, so a(2) = 18 m/s².

3. A particle moves with v = 12t - 3t² m/s for 0 ≤ t ≤ 4. Find its greatest velocity.
   Show answer

   12 m/s. a = 12 - 6t = 0 at t = 2, so v_max = 24 - 12 = 12 m/s.

4. A particle moves with x = t³ - 9t² + 24t metres. Find both times when it is at rest.
   Show answer

   t = 2 s and t = 4 s. v = 3t² - 18t + 24 = 3(t - 2)(t - 4) = 0, so t = 2 or t = 4.

5. For the same particle (x = t³ - 9t² + 24t), find its acceleration at the later time it is at rest.
   Show answer

   6 m/s². a = 6t - 18, so a(4) = 6 m/s².

6. A particle moves with v = 8 - 2e^-t m/s. Find its acceleration when t = ln 2.
   Show answer

   1 m/s². a = 2e^-t. At t = ln 2, e^-t = 1/2, so a = 1 m/s².

7. x = 4t² - t³ for a particle on a line. Why can you not use v = u + at here?
   • The particle starts from rest
   • The displacement is negative at some times
   • The acceleration is not constant
   • v = u + at only works vertically
   Show answer

   The acceleration is not constant. a = 8 - 6t depends on t, and suvat needs constant acceleration.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M2
• Pearson Edexcel IAL Mechanics M2 (WME02) question paper, June 2025: front-cover instructions on g and accuracy
• Pearson Edexcel IAL Mechanics M2 (WME02) examiners' report, June 2025

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