Reference Frames and Relative Motion

Reference frames

A reference frame is the object (and the axes attached to it) from which you measure position and velocity. "The ground," "the train," and "the river water" are all reference frames.

Velocity is always relative to some frame. We write $v_{A/B}$ for "the velocity of A relative to B." Read the slash as "relative to."

Some examples:

• $v_{\text{Mia/train}}$: how fast Mia moves as seen by someone sitting on the train.

• $v_{\text{train/ground}}$: how fast the train moves as seen from the ground.

• $v_{\text{boat/water}}$: the boat's velocity as measured from the water itself.

Adding velocities in one dimension

To change frames, chain the "relative to" labels so the middle ones cancel:

Why does this work? Position in frame C equals position in frame B plus the position of B in frame C. Over the same time interval, the displacements add, so the velocities add. Notice the "B" in the middle: it appears once on the right of the slash and once on the left, like a link in a chain.

Also, swapping the order reverses the sign: $v_{A/B} = -v_{B/A}$. If a runner passes you at $4\ \text{m/s}$ to the right, you move at $4\ \text{m/s}$ to the left relative to the runner.

Relative motion in two dimensions

The equation $v_{A/C} = v_{A/B} + v_{B/C}$ is a vector equation. When the velocities are not along the same line, add them by components:

Then find the magnitude $v = \sqrt{v_x^2 + v_y^2}$ and the angle $\theta = \tan^{-1}(|v_y| / |v_x|)$. Always say which axis the angle is measured from, such as "$37^\circ$ north of east" (measured from east toward north) or "$10^\circ$ east of north" (measured from north toward east). Use the signs of $v_x$ and $v_y$ to pick the quadrant, since a calculator's $\tan^{-1}$ alone can give the wrong one. For example, $(-20,\ 15)$ points up and to the left, so it is $37^\circ$ north of west, which is $143^\circ$ measured from east.

A classic case is a boat crossing a river. The water moves downstream, carrying everything in it. The boat's velocity relative to the bank is the sum of its velocity relative to the water and the water's velocity relative to the bank.

Aiming to cancel the current

What if you want to land directly across from where you start? Then the downstream component of your velocity relative to the bank must be zero. You must aim upstream so that part of your velocity relative to the water cancels the current.

If the swimmer's speed relative to the water is $v_s$, the current is $v_c$, and she aims at an angle $\theta$ upstream from the perpendicular, then:

• Downstream (zero net): $v_s \sin\theta = v_c$

• Across: $v_y = v_s \cos\theta$

The same method works in the air. A plane's airspeed is its speed relative to the air. Wind is the air's velocity relative to the ground, so a plane flying in a crosswind is just a boat in a river. To fly due north over the ground in a wind blowing east, the pilot must head west of north.

Summary