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Acceleration as a function of displacement

Edexcel IAL Mathematics · WME03 Mechanics 3

Sometimes the acceleration depends on where the particle is, not on the time. You cannot integrate a(x) with respect to t, so you use a different form of the acceleration: a = v dv/dx.

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Why a = v dv/dx

By the chain rule, a = dv/dt = (dv/dx)(dx/dt) = v dv/dx. Also v dv/dx = d/dx(½v2). So if a = f(x):

∫v dv = ∫f(x) dx, which gives ½v2 = F(x) + c.

Using it

1. Write v dv/dx = f(x).

2. Separate and integrate: ½v2 = ∫f(x) dx + c.

3. Find c from a known pair (x, v).

4. Answer the question: speed at a position, where the particle comes to rest (v = 0) or where the speed is greatest (a = 0).

Signs and direction

Take the positive direction as the direction x increases. An acceleration "towards O" when x > 0 is negative. When you take a square root of v2, choose the sign that matches the direction of motion.

Worked example

A particle P moves along the x-axis. Its acceleration is (3 - 2x) m/s2 in the direction of x increasing. When x = 0, v = 2 m/s. Find v2 in terms of x, the greatest speed, and where P first comes to rest.

  1. v dv/dx = 3 - 2x, so ½v2 = 3x - x2 + c. At x = 0, ½(4) = c, so c = 2.
  2. v2 = 6x - 2x2 + 4.
  3. Greatest speed when a = 0: x = 1.5, v2 = 8.5, v = 2.92 m/s.
  4. Rest: 2x2 - 6x - 4 = 0, so x2 - 3x - 2 = 0 and x = (3 + sqrt 17)/2 = 3.56 m.

Answer: v2 = 6x - 2x2 + 4; greatest speed 2.92 m/s; rest at x = 3.56 m

Key idea: When a is a function of x, use a = v dv/dx and integrate to get ½v2 = ∫a dx + c.

Check yourself

  1. A particle has acceleration 4x m/s2 and speed 3 m/s when x = 0, moving in the positive direction. Find its speed when x = 2.
    Show answer

    5 m/s. ½v2 = 2x2 + 4.5, so v2 = 4x2 + 9 = 25 and v = 5 m/s.

  2. A particle has acceleration -9x m/s2 and speed 6 m/s at x = 0. Find the value of x > 0 where it comes to rest.
    Show answer

    2 m. ½v2 = -4.5x2 + 18, so v = 0 when x2 = 4, x = 2 m.

  3. A particle moves on the positive x-axis with acceleration -8/x2 m/s2. When x = 1 its speed is 4 m/s, moving away from O. Find its speed when x = 4.
    Show answer

    2 m/s. ½v2 = 8/x + c; at x = 1: 8 = 8 + c, so c = 0 and v2 = 16/x. At x = 4, v2 = 4, v = 2 m/s.

  4. In the worked example (a = 3 - 2x, v = 2 at x = 0), find the greatest speed, to 3 significant figures.
    Show answer

    2.92 m/s. Greatest speed when a = 0, x = 1.5: v2 = 9 - 4.5 + 4 = 8.5, so v = 2.92 m/s.

  5. In the worked example, find the value of x where P first comes to rest, to 3 significant figures.
    Show answer

    3.56 m. v2 = 0 gives x2 - 3x - 2 = 0, x = (3 + sqrt 17)/2 = 3.56 m (the negative root is behind the start).

  6. A particle starts from rest at x = 0 with acceleration e-x m/s2. Its speed approaches a limiting value as x increases. Find it, to 3 significant figures.
    Show answer

    1.41 m/s. ½v2 = -e-x + c, and v = 0 at x = 0 gives c = 1. So v2 = 2(1 - e-x), which tends to 2. Limit = sqrt 2 = 1.41 m/s.

  7. The acceleration of a particle is given as a function of its displacement x. Which form of the acceleration should you use?
    • dv/dt
    • d2v/dx2
    • dx/dt
    • v dv/dx
    Show answer

    v dv/dx. v dv/dx lets you separate the variables v and x.

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