Constant acceleration

Final velocity, displacement and distance travelled under constant acceleration.

Inputs

Starting velocity

Negative for motion in the opposite direction.

Acceleration

Time

Result

Final velocity

4.5

Displacement: 3.375
Distance travelled: 3.375

How it’s calculated

Formula

v = v_0 + a t

s = v_0 t + \tfrac{1}{2} a t^2

t_\text{stop} = -\frac{v_0}{a}, \qquad d = \frac{v_0^2 + v^2}{2|a|} \quad \text{if it reverses, else } d = |s|

With your numbers (converted to SI units)

v = 0 + 3 \times 1.5 = 4.5\,\text{m/s}

s = 0 \times 1.5 + \tfrac{1}{2} \times 3 \times 1.5^2 = 3.375\,\text{m}

d = |s| = 3.375\,\text{m}

What it means

Under constant acceleration, velocity changes linearly with time. Displacement s is where the object ends up relative to its start; distance travelled d is the length of the path. They differ when acceleration opposes the starting velocity for long enough that the object stops and comes back (braking past zero).

Assumptions

Acceleration stays constant for the whole time, including through a stop; real mechanisms taper off as they approach top speed, and brakes stop rather than reverse.

Learn the concepts behind this on resources.simbotics.org · All formulas