Motion can be described from different reference frames. In classical mechanics, if one frame translates relative to another without rotating, their position and velocity descriptions are related by simple vector addition.

Let frame $S'$ have origin at position $\mathbf R(t)$ as measured in frame $S$. If an object has position $\mathbf r'$ in $S'$, then its position in $S$ is

$$\mathbf r=\mathbf R+\mathbf r'.$$

Differentiating with respect to time gives

$$\boxed{\mathbf v=\mathbf V+\mathbf v'},$$

where

$$\mathbf V=\frac{d\mathbf R}{dt}$$

is the velocity of $S'$ relative to $S$.

Equivalently, the velocity of object $A$ relative to object or frame $B$ is

$$\boxed{\mathbf v_{A/B}=\mathbf v_A-\mathbf v_B}.$$

This is the classical rule for relative velocity.

Galilean transformation between inertial frames

If $S'$ moves with constant velocity relative to $S$, then

$$\frac{d\mathbf V}{dt}=\mathbf0.$$

Differentiating the velocity relation gives

$$\mathbf a=\mathbf a'.$$

Thus two nonrotating frames moving at constant relative velocity agree on acceleration. Such frames are both inertial if either one is inertial. Position and velocity differ between them, but Newtonian acceleration does not.

One-dimensional example

A train moves east at $20,\mathrm{m/s}$ relative to the ground. A passenger walks east inside the train at $2,\mathrm{m/s}$ relative to the train. The passenger's ground velocity is

$$v_{P/G}=v_{P/T}+v_{T/G}=2+20=22,\mathrm{m/s}.$$

If the passenger instead walks west at $2,\mathrm{m/s}$ relative to the train,

$$v_{P/G}=-2+20=18,\mathrm{m/s}$$

to the east.

Two-dimensional example: crossing a river

A boat moves north at

$$\mathbf v_{B/W}=(0,4),\mathrm{m/s}$$

relative to the water. The river flows east at

$$\mathbf v_{W/G}=(3,0),\mathrm{m/s}$$

relative to the ground. Therefore

$$\mathbf v_{B/G} =\mathbf v_{B/W}+\mathbf v_{W/G} =(3,4),\mathrm{m/s}.$$

The boat's ground speed is

$$|\mathbf v_{B/G}|=\sqrt{3^2+4^2}=5,\mathrm{m/s}.$$

If the river is $120,\mathrm m$ wide, only the northward component carries the boat across, so

$$t=\frac{120}{4}=30,\mathrm s.$$

During that time the current carries it east by

$$\Delta x=(3)(30)=90,\mathrm m.$$

Relative velocity is therefore a vector relation: the frame motion must be added component by component, not merely added to the speed.