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Velocity as a function of position or time

Edexcel IAL Mathematics · WME03 Mechanics 3

M3 questions often go one step further: once you know v in terms of x, you are asked how long the particle takes to get somewhere, or where it is at a given time. That means solving dx/dt = f(x) by separating the variables.

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dx/dt = f(t)

Integrate directly: x = ∫f(t) dt + c, and find c from the starting position.

dx/dt = f(x)

Separate the variables: ∫ dx/f(x) = ∫ dt. Integrate both sides, find the constant, then rearrange to get x in terms of t (or find t for a given x). Logarithms and inverse trig functions often appear, so use the P3 and P4 integrals.

Exam advice

Pearson's examiners note that candidates often lose marks by ignoring a stated restriction on x, by not simplifying v2 before taking the square root, and by taking the wrong sign of the root. Check that your answer makes sense for the direction of motion.

Worked example

A particle moves along the x-axis so that v = 2(x + 1) m/s. When t = 0, x = 0. Find x in terms of t and the time to reach x = 5.

  1. dx/dt = 2(x + 1), so ∫ dx/(x + 1) = ∫ 2 dt.
  2. ln(x + 1) = 2t + c; at t = 0, x = 0 so c = 0.
  3. x = e2t - 1.
  4. x = 5: e2t = 6, so t = ½ ln 6 = 0.896 s.

Answer: x = e2t - 1; t = 0.896 s

Key idea: dx/dt = f(t): integrate. dx/dt = f(x): separate, ∫dx/f(x) = ∫dt, then use the starting values.

Check yourself

  1. A particle moves so that v = 2(x + 1) m/s, with x = 0 at t = 0. Find the time it takes to reach x = 5, to 3 significant figures.
    Show answer

    0.896 s. ln(x + 1) = 2t, so t = ½ ln 6 = 0.896 s.

  2. A particle moves so that v = 4/(x + 2) m/s, with x = 0 when t = 0. Find the time it takes to reach x = 4.
    Show answer

    4 s. ∫(x + 2) dx = ∫4 dt gives x2/2 + 2x = 4t (constant 0). At x = 4: 8 + 8 = 4t, t = 4 s.

  3. A particle moves so that v = 5e-x m/s, with x = 0 when t = 0. Find x when t = 2, to 3 significant figures.
    Show answer

    2.40 m. ∫ex dx = ∫5 dt gives ex = 5t + 1. At t = 2, ex = 11, x = ln 11 = 2.40 m.

  4. A particle has velocity (6t - 3t2) m/s and is at x = 2 m when t = 0. Find x when t = 2.
    Show answer

    6 m. x = 3t2 - t3 + 2, so x(2) = 12 - 8 + 2 = 6 m.

  5. A particle moves with v = 2 sqrt(9 - x2) m/s in the positive direction, with x = 0 when t = 0. Find x when t = 0.5, to 3 significant figures.
    Show answer

    2.52 m. ∫ dx/sqrt(9 - x2) = ∫2 dt gives arcsin(x/3) = 2t, so x = 3 sin 2t. At t = 0.5, x = 3 sin 1 = 2.52 m.

  6. Velocity is given as v = f(x). Which statement shows how to find the time taken to move from x = p to x = q?
    • t = ∫ from p to q of 1/f(x) dx
    • t = ∫ from p to q of f(x) dx
    • t = f(q) - f(p)
    • t = (q - p)/f(p)
    Show answer

    t = ∫ from p to q of 1/f(x) dx. dt/dx = 1/v = 1/f(x), so integrate 1/f(x).

  7. From v2 = 4(9 - x2) you need v for a particle moving towards O from x = 2. Which is correct?
    • v = 2 sqrt(9 - x2)
    • v = 2(3 - x)
    • v = -2 sqrt(9 - x2)
    • v = 4(9 - x2)
    Show answer

    v = -2 sqrt(9 - x2). Moving towards O from positive x means x is decreasing, so v is negative.

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