A coordinate system in which an experimenter (possibly with the assistance of helpers) makes position and time measurements of physical events is called a reference frame. In Figure 4.17, Amy and Bill each had their own reference frame (where they were at rest) in which they measured Carlos’s velocity.

More generally, FIGURE 4.18 shows two reference frames, A and B, and an object C. It is assumed that the reference frames are moving with respect to each other. At this instant of time, the position vector of C in reference frame A is $\vec{r}{}_{\text{CA}}$, meaning “the position of C relative to the origin of frame A.” Similarly, $\vec{r}{}_{\text{CB}}$ is the position vector of C in reference frame B. Using vector addition, you can see that

FIGURE 4.18 Two reference frames.

where $\vec{r}{}_{\text{AB}}$ locates the origin of A relative to the origin of B.

In general, object C is moving relative to both reference frames. To find its velocity in each reference frame, take the time derivative of Equation 4.18:

By definition, $d\vec{r}\text{/}dt$ is a velocity. The first derivative in Equation 4.19 is $\vec{v}{}_{\text{CB}}$, the velocity of C relative to B. Similarly, the second is the velocity of C relative to A, $\vec{v}{}_{\text{CA}}$. The last derivative is slightly different because it doesn’t refer to object C. Instead, this is the velocity $\vec{v}{}_{\text{AB}}$ of reference frame A relative to reference frame B. As we noted in one dimension, $\vec{v}{}_{\text{AB}}=-\vec{v}{}_{\text{BA}}$.

Writing Equation 4.19 in terms of velocities, we have

This relationship between velocities in different reference frames was recognized by Galileo in his pioneering studies of motion, hence it is known as the Galilean transformation of velocity. If you know an object’s velocity in one reference frame, you can transform it into the velocity that would be measured in a different reference frame. Just as in one dimension, the velocity of C relative to B is the velocity of C relative to A plus the velocity of A relative to B, but you must add the velocities as vectors for two-dimensional motion.

As we’ve seen, the Galilean velocity transformation is pretty much common sense for one-dimensional motion. The real usefulness appears when an object travels in a medium moving with respect to the earth. For example, a boat moves relative to the water. What is the boat’s net motion if the water is a flowing river? Airplanes fly relative to the air, but the air at high altitudes often flows at high speed. Navigation of boats and planes requires knowing both the motion of the vessel in the medium and the motion of the medium relative to the earth.

EXAMPLE 4.7 Flying to Cleveland I

Cleveland is 300 miles east of Chicago. A plane leaves Chicago flying due east at 500 mph. The pilot forgot to check the weather and doesn’t know that the wind is blowing to the south at 50 mph. What is the plane’s ground speed? Where is the plane 0.60 h later, when the pilot expects to land in Cleveland?

MODEL Establish a coordinate system with the x-axis pointing east and the y-axis north. The plane P flies in the air, so its velocity relative to the air A is $\vec{v}{}_{\text{PA}}=500î\text{m}\text{p}\text{h}$. Meanwhile, the air is moving relative to the ground G at $\vec{v}{}_{\text{AG}}=-50ĵ\text{m}\text{p}\text{h}$.

SOLVE The velocity equation $\vec{v}{}_{\text{PG}}=\vec{v}{}_{\text{PA}}+\vec{v}{}_{\text{AG}}$ is a vector-addition equation. FIGURE 4.19 shows graphically what happens. Although the nose of the plane points east, the wind carries the plane in a direction somewhat south of east. The plane’s velocity relative to the ground is

$$\vec{v}{}_{\text{PG}}=\vec{v}{}_{\text{PA}}+\vec{v}{}_{\text{AG}}=(500î-50ĵ)\text{m}\text{p}\text{h}$$

FIGURE 4.19 The wind causes a plane flying due east in the air to move to the southeast relative to the ground.

FIGURE 4.19 Full Alternative Text

The plane’s ground speed is

$$v=\sqrt{{\left(v_x\right)}_{\text{PG}}{\text{}}^2+{\left(v_y\right)}_{\text{PG}}{}^2}=502\text{ mph}$$

After flying for 0.60 h at this velocity, the plane’s location (relative to Chicago) is

$$\begin{array}{l}x={\left(v_x\right)}_{\text{PG}}t=(500\text{ mph})(0.60\text{ h})=300\text{ mi}\hfill \\ y={\left(v_y\right)}_{\text{PG}}t=(-50\text{ mph})(0.60\text{ h})=-30\text{ mi}\hfill \end{array}$$

The plane is $30\text{mi}$ due south of Cleveland! Although the pilot thought he was flying to the east, his actual heading has been ${\text{tan}}^{-1}(50\text{mph}\text{/}500\text{m}\text{p}\text{h})={\text{tan}}^{-1}\left(0.10\right)=5.71\text{°}$ south of east.

EXAMPLE 4.8 Flying to Cleveland II

A wiser pilot flying from Chicago to Cleveland on the same day plots a course that will take her directly to Cleveland. In which direction does she fly the plane? How long does it take to reach Cleveland?

MODEL Establish a coordinate system with the x-axis pointing east and the y-axis north. The air is moving relative to the ground at $\vec{v}{}_{\text{AG}}=-50ĵ\text{m}\text{p}\text{h}$.

SOLVE The objective of navigation is to move between two points on the earth’s surface. The wiser pilot, who knows that the wind will affect her plane, draws the vector picture of FIGURE 4.20. She sees that she’ll need ${\left(v_y\right)}_{\text{PG}}=0$ in order to fly due east to Cleveland. This will require turning the nose of the plane at an angle $\theta$ north of east, making $\vec{v}{}_{\text{PA}}=(500\text{cos}\theta î+500\text{sin}\theta ĵ)\text{m}\text{p}\text{h}$.

FIGURE 4.20 To travel due east in a south wind, a pilot has to point the plane somewhat to the northeast.

FIGURE 4.20 Full Alternative Text

The velocity equation is $\vec{v}{}_{\text{PG}}=\vec{v}{}_{\text{PA}}+\vec{v}{}_{\text{AG}}$. The desired heading is found from setting the y-component of this equation to zero:

$$\begin{array}{l}{\left(v_y\right)}_{\text{PG}}={\left(v_y\right)}_{\text{PA}}+{\left(v_y\right)}_{\text{AG}}=(500\text{ sin }\theta -50)\text{ mph}=0\text{ mph}\hfill \\ \theta ={\text{sin}}^{-1}\left({\displaystyle \frac{50\text{ mph}}{500\text{ mph}}}\right)=5.74\text{°}\hfill \end{array}$$

The plane’s velocity relative to the ground is then $\vec{v}{}_{\text{PG}}=$ $500\text{cos}5.74\text{°}î\text{mph}=497î\text{mph}$. This is slightly slower than the speed relative to the air. The time needed to fly to Cleveland at this speed is

$$t={\displaystyle \frac{300\text{mi}}{497\text{mph}}}=0.604\text{h}$$

It takes $0.004\text{h}=14\text{s}$ longer to reach Cleveland than it would on a day without wind.

REVIEW A boat crossing a river or an ocean current faces the same difficulties. These are exactly the kinds of calculations performed by pilots of boats and planes as part of navigation.