Kinematic equations for uniformly accelerated motion

For constant acceleration a, with initial velocity v₀ (at t = 0) and velocity v at time t, three relations connect x, v, a and t: v = v₀ + at, x = v₀t + ½at², and v² = v₀² + 2ax.

Here x is the displacement from the starting position, not the distance travelled; if the body turns around, find the turning point first.

The average velocity over the interval is (v₀ + v)/2 for constant acceleration only, which gives x = (v₀ + v)t/2.

The equations can be obtained graphically (area under the v-t graph) or by integrating a = dv/dt and v = dx/dt with constant a.

Choose the positive direction first, then give every vector quantity (v₀, v, a, x) its sign from that choice.

Convert all speeds to m s⁻¹ before substituting: multiply km/h by 5/18.

For a body that accelerates and then decelerates from rest to rest, the same peak speed links both phases; the v-t graph is a triangle whose area is the total displacement.