Uniform motion

Uniform motion is motion at an even rate – that is, at constant velocity. Since the velocity is constant, its change is zero, so the acceleration is zero:

\[v_x = \mathrm{const}\]

\[a_x = \frac{\mathrm{d}v_x}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \mathrm{const} = 0\]

A constant velocity means that the position changes at an even rate, and the length of the interval does not influence this rate. Consequently, we do not have apply differential calculus, but we can take a finite time interval \(\Delta t\):

\[v_x = \frac{\Delta x}{\Delta t} \Rightarrow \Delta x = v_x \Delta t.\]

The displacement \(\Delta x\) is the final position \(x\) minus the initial position \(x_0,\) whilst the time elapsed \(\Delta t\) is the final time \(t\) minus the initial time \(t_0\):

\[x - x_0 = v_x \left(t - t_0\right) \Rightarrow x = x_0 + v_x t - v_x t_0.\]

We can assume without loss of generality that we started our stopwatch when the motion began: \(t_0 = 0.\) With this, our formula becomes simpler: the position \(x\) at any given time \(t\) is

\[x(t) = x_0 + v_x t.\]