# 3.5 Addition of velocities  (Page 3/12)

 Page 3 / 12

Solution

Because ${\mathbf{\text{v}}}_{\text{tot}}$ is the vector sum of the ${\mathbf{\text{v}}}_{\text{w}}$ and ${\mathbf{\text{v}}}_{\text{p}}$ , its x - and y -components are the sums of the x - and y -components of the wind and plane velocities. Note that the plane only has vertical component of velocity so ${v}_{px}=0$ and ${v}_{py}={v}_{\text{p}}$ . That is,

${v}_{\text{tot}x}={v}_{\text{w}x}$

and

${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}\text{.}$

We can use the first of these two equations to find ${v}_{\text{w}x}$ :

${v}_{\text{w}y}={v}_{\text{tot}x}={v}_{\text{tot}}\text{cos 110º}\text{.}$

Because ${v}_{\text{tot}}=\text{38}\text{.}0 m/\text{s}$ and $\text{cos 110º}=–0.342$ we have

${v}_{\text{w}y}=\left(\text{38.0 m/s}\right)\left(\text{–0.342}\right)=\text{–13 m/s.}$

The minus sign indicates motion west which is consistent with the diagram.

Now, to find ${v}_{\text{w}\text{y}}$ we note that

${v}_{\text{tot}y}={v}_{\text{w}y}+{v}_{\text{p}}$

Here ; thus,

${v}_{\text{w}y}=\left(\text{38}\text{.}0 m/s\right)\left(0\text{.}\text{940}\right)-\text{45}\text{.}0 m/s=-9\text{.}\text{29 m/s.}$

This minus sign indicates motion south which is consistent with the diagram.

Now that the perpendicular components of the wind velocity ${v}_{\text{w}x}$ and ${v}_{\text{w}y}$ are known, we can find the magnitude and direction of ${\mathbf{\text{v}}}_{\text{w}}$ . First, the magnitude is

$\begin{array}{lll}{v}_{\text{w}}& =& \sqrt{{v}_{\text{w}x}^{2}+{v}_{\text{w}y}^{2}}\\ & =& \sqrt{\left(-\text{13}\text{.}0 m/s{\right)}^{2}+\left(-9\text{.}\text{29 m/s}{\right)}^{2}}\end{array}$

so that

${v}_{\text{w}}=\text{16}\text{.}0 m/s\text{.}$

The direction is:

$\theta ={\text{tan}}^{-1}\left({v}_{\text{w}y}/{v}_{\text{w}x}\right)={\text{tan}}^{-1}\left(-9\text{.}\text{29}/-\text{13}\text{.}0\right)$

giving

$\theta =\text{35}\text{.}6º\text{.}$

Discussion

The wind’s speed and direction are consistent with the significant effect the wind has on the total velocity of the plane, as seen in [link] . Because the plane is fighting a strong combination of crosswind and head-wind, it ends up with a total velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.

Note that in both of the last two examples, we were able to make the mathematics easier by choosing a coordinate system with one axis parallel to one of the velocities. We will repeatedly find that choosing an appropriate coordinate system makes problem solving easier. For example, in projectile motion we always use a coordinate system with one axis parallel to gravity.

## Relative velocities and classical relativity

When adding velocities, we have been careful to specify that the velocity is relative to some reference frame . These velocities are called relative velocities . For example, the velocity of an airplane relative to an air mass is different from its velocity relative to the ground. Both are quite different from the velocity of an airplane relative to its passengers (which should be close to zero). Relative velocities are one aspect of relativity    , which is defined to be the study of how different observers moving relative to each other measure the same phenomenon.

Nearly everyone has heard of relativity and immediately associates it with Albert Einstein (1879–1955), the greatest physicist of the 20th century. Einstein revolutionized our view of nature with his modern theory of relativity, which we shall study in later chapters. The relative velocities in this section are actually aspects of classical relativity, first discussed correctly by Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of light—that is, less than . Most things we encounter in daily life move slower than this speed.

Let us consider an example of what two different observers see in a situation analyzed long ago by Galileo. Suppose a sailor at the top of a mast on a moving ship drops his binoculars. Where will it hit the deck? Will it hit at the base of the mast, or will it hit behind the mast because the ship is moving forward? The answer is that if air resistance is negligible, the binoculars will hit at the base of the mast at a point directly below its point of release. Now let us consider what two different observers see when the binoculars drop. One observer is on the ship and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, and so he sees them fall straight down the mast. (See [link] .) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance forward while the binoculars are falling. This observer sees the curved path shown in [link] . Although the paths look different to the different observers, each sees the same result—the binoculars hit at the base of the mast and not behind it. To get the correct description, it is crucial to correctly specify the velocities relative to the observer.

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º20.0º with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate an
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