All subjects · Physics · WPH11 (Unit 1)

Equations of uniformly accelerated motion

When acceleration is constant, five quantities are linked by four equations. They are the backbone of Unit 1 mechanics.

The five quantities

Displacement, velocity and acceleration are vectors, so their signs matter.

The four equations

Write down the three quantities you know and the one you want, then choose the equation that leaves out the fifth. They only work for uniform (constant) acceleration in a straight line.

Free fall and signs

Near the Earth's surface, a freely falling object has acceleration g = 9.81 m/s2 downwards (the value on the Pearson data list). Pick a positive direction and keep it. If up is positive, a = -9.81 m/s2 for anything moving freely, whether it is going up or coming down. At the top of its flight, an object thrown straight up has v = 0 for an instant.

Worked example

A ball is thrown vertically upwards at 12 m/s. Find the maximum height it reaches.

  1. Take up as positive: u = 12 m/s, v = 0 at the top, a = -9.81 m/s2, want s.
  2. v2 = u2 + 2as gives 0 = 144 + 2(-9.81)s
  3. s = 144 / 19.62 = 7.34 m

Answer: Maximum height = 7.3 m (2 s.f.)

Key idea: List s, u, v, a, t; pick the equation without the unwanted quantity; one positive direction; g = 9.81 m/s2.

Check yourself

  1. A car moving at 5.0 m/s accelerates at 1.5 m/s2 for 8.0 s. Find its final velocity in m/s.
    Show answer

    17 m/s. v = u + at = 5.0 + 1.5 x 8.0 = 17 m/s.

  2. A cyclist starts from rest and accelerates uniformly at 2.4 m/s2 for 5.0 s. How far does she travel, in m?
    Show answer

    30 m. s = ut + (1/2)at2 = 0 + 0.5 x 2.4 x 25 = 30 m.

  3. A train travelling at 30 m/s brakes uniformly to rest over 450 m. Find the size of its deceleration in m/s2.
    Show answer

    1 m/s2. 0 = 302 + 2a(450), so a = -900/900 = -1.0 m/s2. The deceleration is 1.0 m/s2.

  4. A stone is dropped from rest and falls freely for 2.0 s. How far does it fall, in m? Give 3 s.f.
    Show answer

    19.6 m. s = (1/2)gt2 = 0.5 x 9.81 x 4.0 = 19.6 m.

  5. An object is dropped from rest from a height of 20 m. Find its speed just before it hits the ground, in m/s. Ignore air resistance and give 3 s.f.
    Show answer

    19.8 m/s. v2 = 0 + 2 x 9.81 x 20 = 392.4, so v = 19.8 m/s.

  6. A trolley speeds up uniformly from 4.0 m/s to 10 m/s in 6.0 s. How far does it travel, in m?
    Show answer

    42 m. s = (1/2)(u + v)t = 0.5 x 14 x 6.0 = 42 m.

  7. You know u, a and s and want v. Which equation should you use?
    • v = u + at
    • s = ut + (1/2)at2
    • s = (1/2)(u + v)t
    • v2 = u2 + 2as
    Show answer

    v2 = u2 + 2as. v2 = u2 + 2as contains u, a, s and v and leaves out t, which you do not know.

  8. A ball is thrown straight up. Taking upwards as positive, what is its acceleration while it is still rising?
    • +9.81 m/s2
    • 0 m/s2
    • -9.81 m/s2
    • It changes sign at the top
    Show answer

    -9.81 m/s2. Gravity always pulls down, so a = -9.81 m/s2 throughout the flight, including at the top.

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Last updated 7 November 2026. Spotted a mistake? Tell us.