# 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.

Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
hour glass, pendulum clock, atomic clock?
S.M
tnks
David
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
the topic is kinematics
David
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Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
is acceleration a fundamental unit.
no it is derived
Abdul
no
Nisha
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David
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Emmanuel
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how about a conceptual framework can you simplify for me? needed please
Villaflor
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Afzal
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Villaflor
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S.M
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S.M
Physics
Beatriz
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David
equations of kinematics for constant acceleration
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
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Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
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who can help me with my problem about acceleration?
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
no
Emmanuel
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Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
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v=d/t
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Villaflor
Villaflor
v=d/t
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Nuru
A hydrometer of mass 0.15kg and uniform cross sectional area of 0.0025m2 displaced in water of density 1000kg/m3.what depth will the hydrometer sink
Lila
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Darshik
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aways
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aways
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