Equations of uniformly accelerated motion

When the acceleration is constant, three equations link the five quantities v₀ (starting velocity), v, a, t and x:

• v = v₀ + at (from the meaning of acceleration).
• x – x₀ = v₀t + ½at² (the area under the v–t graph: a rectangle plus a triangle).
• v² = v₀² + 2a(x – x₀) (put the first into the second).

Use the one equation that leaves out the quantity you are not given and do not need. They only work for constant acceleration.

Before using them, choose a positive direction and give every quantity its sign. For a ball thrown upward with up as positive, a = –g = –9.8 m s⁻² the whole time, on the way up and on the way down. At the top the velocity is zero. The displacement can come out negative: that just means the body ends up below its starting point.

Example: a ball is thrown up at 19.6 m s⁻¹ from the top of a tower and hits the ground 6 s later. With up as positive, the displacement is –h = 19.6 × 6 – ½ × 9.8 × 6² = 117.6 – 176.4, so the tower is 58.8 m high.

Something that drops off a moving body starts with that body's velocity. A packet let go from a balloon rising at 5 m s⁻¹ first moves up at 5 m s⁻¹, slows, and only then falls.

Some useful results:

• From rest, a falling body covers distances in equal time intervals in the ratio 1 : 3 : 5 : 7 … (Galileo's law of odd numbers).
• The stopping distance is d = v₀²/(2a): double the speed and the stopping distance becomes four times as long. This is why speed limits matter.
• Your reaction time can be measured with a falling ruler: if it drops d before you catch it, t = √(2d/g).