Stopping Distance

Written by Jerry Ratzlaff on . Posted in Classical Mechanics

formula

\(d = \frac { v^2 } { 2 \mu g }   \)

\(d = k v^2   \)

Where:

\(d\) = stopping distance

\(v\) = velocity

\(\mu\) = friction coefficient

\(g\) = gravitational acceleration

\(k\) = constant

Solve for:

\(k = \frac { d } { v^2 }   \)