What Is the Angle Theta With Respect to North

Learning Objectives

By the end of this section, y'all will be able to:

• Apply principles of vector addition to make up one's mind relative velocity.
• Explain the significance of the observer in the measurement of velocity.

Relative Velocity

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a potent crosswind, you tin sometimes see that the plane is not moving in the direction in which information technology is pointed, as illustrated in Figure ii. The plane is moving directly ahead relative to the air, merely the motion of the air mass relative to the ground carries it sideways.

Figure 1. A boat trying to caput straight across a river will actually motion diagonally relative to the shore every bit shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

Figure 2. An airplane heading straight due north is instead carried to the westward and slowed down past wind. The plane does non move relative to the ground in the direction it points; rather, it moves in the management of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such every bit a river) and that medium has a velocity relative to an observer on solid basis. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are just two of many in which it is useful to add together velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means. How do nosotros add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Vector Improver and Subtraction: Graphical Methods and Vector Improver and Subtraction: Analytical Methods employ to the addition of velocities, just as they do for any other vectors. In one-dimensional motility, the addition of velocities is elementary—they add similar ordinary numbers. For example, if a field hockey player is moving at five chiliad/southward straight toward the goal and drives the ball in the aforementioned management with a velocity of 30 m/s relative to her trunk, then the velocity of the ball is 35 one thousand/south relative to the stationary, profusely sweating goalkeeper standing in front of the goal. In two-dimensional move, either graphical or belittling techniques tin be used to add velocities. We will concentrate on belittling techniques. The post-obit equations give the relationships between the magnitude and management of velocity (v and θ) and its components (five ₁₀ and v _y ) forth the x – and y -axes of an appropriately chosen coordinate system:

v ₁₀ = v cosθ

5 _y = v sinθ

[latex]five=\sqrt{{{5}_{x}}^{2}+{{five}_{y}}^{2}}[/latex]

θ = tan ^{− 1} (5 _y /v _x ).

Figure 3. The velocity, five , of an object traveling at an angle θ to the horizontal axis is the sum of component vectors 5 _ten and v _y.

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to observe the magnitude and direction of velocity when its components are known.

Take-Home Experiment: Relative Velocity of a Boat

Fill up a bathtub one-half-full of water. Take a toy boat or some other object that floats in h2o. Unplug the drain so water starts to drain. Try pushing the gunkhole from ane side of the tub to the other and perpendicular to the flow of water. Which way do you demand to button the boat then that it ends up immediately opposite? Compare the directions of the flow of water, heading of the boat, and actual velocity of the boat.

Instance 1. Adding Velocities: A Boat on a River

Figure 4. A boat attempts to travel straight across a river at a speed 0.75 chiliad/s. The electric current in the river, however, flows at a speed of i.20 m/s to the right. What is the total displacement of the boat relative to the shore?

Refer to Figure 4 which shows a boat trying to go straight across the river. Allow the states calculate the magnitude and direction of the boat's velocity relative to an observer on the shore, v_tot. The velocity of the gunkhole, five_boat, is 0.75 one thousand/due south in the y -management relative to the river and the velocity of the river, v_river, is 1.20 chiliad/due south to the right.

Strategy

We start by choosing a coordinate organisation with its x-axis parallel to the velocity of the river, every bit shown in Figure four. Because the boat is directed straight toward the other shore, its velocity relative to the water is parallel to the y-centrality and perpendicular to the velocity of the river. Thus, we tin can add the two velocities by using the equations [latex]{5}_{\text{tot}}=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{two}}[/latex] and θ= tan⁻¹(v _y /five _x ) directly.

Solution

The magnitude of the total velocity is

[latex]{v}_{\text{tot}}=\sqrt{{{v}_{x}}^{ii}+{{five}_{y}}^{ii}}[/latex]

where

v ₁₀ = v _river = 1.20 g/s

and

v _y =five _gunkhole= 0.750 thou/south.

Thus,

[latex]{v}_{\text{tot}}=\sqrt{({1.20}\text{ thou/south})^{2} + ({0.750}\text{ m/southward})^{2}}[/latex]

yielding

v _tot= 1.42 chiliad/s.

The management of the total velocity θ is given by:

θ= tan⁻¹(5 _y /5 _x ) = tan⁻¹(0.750/1.twenty).

This equation gives

θ= 32.0º.

Discussion

Both the magnitude v and the management θ of the total velocity are consequent with Effigy four. Note that because the velocity of the river is large compared with the velocity of the boat, it is swept apace downstream. This result is evidenced by the minor bending (only 32.0º) the total velocity has relative to the riverbank.

Example ii. Calculating Velocity: Current of air Velocity Causes an Airplane to Drift

Calculate the air current velocity for the state of affairs shown in Figure 5. The plane is known to be moving at 45.0 grand/s due n relative to the air mass, while its velocity relative to the ground (its total velocity) is 38.0 thou/s in a direction 20.0º w of north.

Figure v. An airplane is known to exist heading north at 45.0 m/s, though its velocity relative to the basis is 38.0 m/s at an angle west of due north. What is the speed and direction of the wind?

Strategy

In this problem, somewhat unlike from the previous instance, we know the total velocity v_tot and that it is the sum of two other velocities, v_w (the wind) and v_p (the aeroplane relative to the air mass). The quantity 5_p is known, and we are asked to find v_w. None of the velocities are perpendicular, but it is possible to find their components along a common set of perpendicular axes. If we tin can discover the components of v_w, then we can combine them to solve for its magnitude and direction. Equally shown in Figure v, we choose a coordinate organisation with its x-axis due east and its y-axis due north (parallel to v_p). (You may wish to wait back at the discussion of the addition of vectors using perpendicular components in Vector Improver and Subtraction: Analytical Methods.)

Solution

Because v_tot is the vector sum of the v_west and five_p, its x– and y-components are the sums of the x– and y-components of the wind and plane velocities. Note that the aeroplane only has vertical component of velocity so 5 _px = 0 and v _py =5 _p. That is,

five _{tot x} = 5 _{w x}

and

v _toty =v _wx +v _p.

Nosotros tin can use the starting time of these two equations to find v _wx :

v _westx =v _totx =v _totcos 110º.

Because 5 _tot = 38.0 m/s and cos 110º=–0.342 nosotros have

five _westx = (38.0 m/s)(–0.342) = –13.0 m/south.

The minus sign indicates move west which is consequent with the diagram. Now, to notice 5 _wy we note that

v _{tot y} = v _{w 10} + five _p

Hither v _toty = v _totsin 110º; thus,

v _westwardy = (38.0 1000/southward)(0.940)−45.0 m/s = −9.29 thou/southward.

This minus sign indicates motion south which is consistent with the diagram. Now that the perpendicular components of the air current velocity v _westx and 5 _wy are known, nosotros can discover the magnitude and direction of v_westward. Kickoff, the magnitude is

[latex]\brainstorm{array}{c}{{5}_{due west}}\hfill=\hfill\sqrt{{{v}_{wx}}^{ii}+{{v}_{wy}}^{2}} \\\hfill=\hfill\sqrt{({-13.0}\text{ m/south})^{2} + ({-9.29}\text{ m/s}^{2})}\end{array}[/latex]

and so that

five _{due west}= xvi.0 m/southward.

The direction is:

θ = tan^{− i} (v _wy /v _wx ) = tan^{− 1}(−ix.29/−xiii.0)

giving

θ= 35.6º.

Word

The wind's speed and direction are consistent with the significant effect the wind has on the full velocity of the plane, as seen in Effigy five. Because the plane is fighting a strong combination of crosswind and head-wind, information technology ends upward with a full velocity significantly less than its velocity relative to the air mass as well as heading in a different direction.

Notation that in both of the last 2 examples, nosotros were able to make the mathematics easier by choosing a coordinate system with 1 axis parallel to one of the velocities. We volition repeatedly find that choosing an appropriate coordinate system makes trouble 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 calculation velocities, nosotros 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 plane 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 shut to zero). Relative velocities are one attribute of relativity, which is divers to be the study of how different observers moving relative to each other mensurate the same phenomenon.

Nearly anybody 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 modernistic theory of relativity, which nosotros shall study in later on chapters. The relative velocities in this section are really aspects of classical relativity, starting time discussed correctly past Galileo and Isaac Newton. Classical relativity is limited to situations where speeds are less than about 1% of the speed of low-cal—that is, less than iii,000 km/south. Almost things we encounter in daily life move slower than this speed.

Let us consider an example of what two dissimilar observers see in a situation analyzed long ago past Galileo. Suppose a sailor at the tiptop 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 forrard? The answer is that if air resistance is negligible, the binoculars volition hitting at the base of the mast at a point directly below its indicate of release. Now let u.s. consider what 2 dissimilar observers see when the binoculars drop. One observer is on the transport and the other on shore. The binoculars have no horizontal velocity relative to the observer on the ship, then he sees them fall direct down the mast. (See Effigy half-dozen.) To the observer on shore, the binoculars and the ship have the same horizontal velocity, so both move the same distance frontward while the binoculars are falling. This observer sees the curved path shown in Figure 6. Although the paths look different to the dissimilar observers, each sees the aforementioned result—the binoculars hit at the base of operations of the mast and non behind it. To get the correct description, information technology is crucial to correctly specify the velocities relative to the observer.

Effigy 6. Classical relativity. The same movement as viewed by two different observers. An observer on the moving ship sees the binoculars dropped from the superlative of its mast fall straight down. An observer on shore sees the binoculars take the curved path, moving forwards with the ship. Both observers see the binoculars strike the deck at the base of the mast. The initial horizontal velocity is dissimilar relative to the two observers. (The ship is shown moving rather fast to emphasize the effect.)

Instance three. Computing Relative Velocity: An Airline Rider Drops a Coin

An airline passenger drops a coin while the plane is moving at 260 m/due south. What is the velocity of the money when it strikes the floor 1.50 m below its betoken of release: (a) Measured relative to the plane? (b) Measured relative to the Earth?

Figure vii. The motility of a money dropped within an airplane as viewed by two different observers. (a) An observer in the plane sees the coin autumn directly down. (b) An observer on the ground sees the coin move almost horizontally.

Strategy

Both problems can be solved with the techniques for falling objects and projectiles. In role (a), the initial velocity of the coin is cypher relative to the plane, and so the motility is that of a falling object (one-dimensional). In part (b), the initial velocity is 260 m/due south horizontal relative to the World and gravity is vertical, so this motility is a projectile movement. In both parts, it is best to use a coordinate system with vertical and horizontal axes.

Solution for (a)

Using the given information, nosotros note that the initial velocity and position are naught, and the terminal position is 1.50 m. The terminal velocity tin be plant using the equation:

v _y ² = v _0y ² −2one thousand(y−y ₀).

Substituting known values into the equation, we get

[latex]{{v}_{y}}^{ii}={0}^{two}-2\left(nine\text{.}\text{lxxx}{\text{m/due south}}^{two}\right)\left(-1\text{.}\text{50}\text{chiliad}-0 grand\correct)=\text{29}\text{.}4{\text{m}}^{2}{\text{/southward}}^{2}[/latex]

yielding

five _y = −5.42 m/s.

We know that the square root of 29.4 has two roots: 5.42 and -5.42. Nosotros cull the negative root because we know that the velocity is directed downward, and nosotros have defined the positive management to exist upward. There is no initial horizontal velocity relative to the airplane and no horizontal acceleration, then the motion is straight down relative to the plane.

Solution for (b)

Because the initial vertical velocity is zero relative to the footing and vertical movement is independent of horizontal motion, the final vertical velocity for the coin relative to the footing is v_y = -5.42 grand/s, the same equally establish in part (a). In contrast to part (a), in that location now is a horizontal component of the velocity. Yet, since in that location is no horizontal acceleration, the initial and final horizontal velocities are the aforementioned and v _x =260 1000/south. The x– and y-components of velocity can exist combined to find the magnitude of the final velocity:

[latex]five=\sqrt{{{5}_{x}}^{2}+{{v}_{y}}^{2}}[/latex].

Thus,

[latex]v=\sqrt{({260\text{ m/s})}^{two}+({-5.42\text{ m/due south}})^{ii}}[/latex]

yielding

5= 260.06 chiliad/s.

The direction is given past:

θ = tan⁻¹(5 _y /v _x ) = tan^−ane(−5.42/260)

then that

θ= tan⁻ⁱ(−0.0208)=−1.19º.

Discussion

In part (a), the final velocity relative to the plane is the same as information technology would be if the coin were dropped from rest on the Earth and fell 1.50 m. This event fits our experience; objects in a plane fall the same mode when the plane is flying horizontally equally when it is at rest on the basis. This result is as well true in moving cars. In part (b), an observer on the ground sees a much different motion for the coin. The aeroplane is moving so fast horizontally to begin with that its final velocity is barely greater than the initial velocity. Again, we see that in ii dimensions, vectors do non add like ordinary numbers—the final velocity v in part (b) is non (260 – five.42) m/southward; rather, it is 260.06 yard/southward. The velocity's magnitude had to be calculated to five digits to see any difference from that of the airplane. The motions as seen by different observers (one in the aeroplane and one on the ground) in this example are analogous to those discussed for the binoculars dropped from the mast of a moving ship, except that the velocity of the plane is much larger, so that the 2 observers see very dissimilar paths. (Come across Effigy seven.) In addition, both observers see the coin fall one.l one thousand vertically, just the one on the ground also sees information technology motion forwards 144 m (this calculation is left for the reader). Thus, one observer sees a vertical path, the other a about horizontal path.

Making Connections: Relativity and Einstein

Because Einstein was able to conspicuously define how measurements are made (some involve light) and considering the speed of light is the same for all observers, the outcomes are spectacularly unexpected. Fourth dimension varies with observer, energy is stored every bit increased mass, and more surprises await.

PhET Explorations: Motion in 2nd

Try the new "Ladybug Motion 2D" simulation for the latest updated version. Learn about position, velocity, and dispatch vectors. Motility the brawl with the mouse or permit the simulation move the ball in 4 types of motion (2 types of linear, simple harmonic, circle).

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Summary

• Velocities in 2 dimensions are added using the same analytical vector techniques, which are rewritten equally

  v _x = v cosθ

  v _y = v sinθ

  [latex]five=\sqrt{{{5}_{x}}^{two}+{{5}_{y}}^{two}}[/latex]

  θ = tan ^{− 1} (5 _y /v ₁₀ ).

• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
• Relativity is the written report of how different observers measure the aforementioned phenomenon, specially when the observers move relative to ane another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).

Conceptual Questions

1. What frame or frames of reference exercise you instinctively use when driving a automobile? When flying in a commercial jet plane?
2. A basketball player dribbling downwardly the court usually keeps his eyes fixed on the players around him. He is moving fast. Why doesn't he demand to keep his eyes on the ball?
3. If someone is riding in the back of a pickup truck and throws a softball straight backward, is information technology possible for the ball to fall straight down as viewed by a person continuing at the side of the road? Under what condition would this occur? How would the move of the brawl appear to the person who threw it?
4. The hat of a jogger running at abiding velocity falls off the back of his head. Depict a sketch showing the path of the hat in the jogger's frame of reference. Draw its path every bit viewed by a stationary observer.
5. A clod of dirt falls from the bed of a moving truck. Information technology strikes the ground directly below the end of the truck. What is the management of its velocity relative to the truck just before it hits? Is this the aforementioned as the direction of its velocity relative to basis just earlier it hits? Explain your answers.

Issues & Exercises

ane. Bryan Allen pedaled a man-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45º due south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his move relative to the Earth. What was his average velocity relative to the air? (c) What was his full displacement relative to the air mass?

2. A seagull flies at a velocity of 9.00 m/south straight into the air current. (a) If it takes the bird xx.0 min to travel vi.00 km relative to the Earth, what is the velocity of the air current? (b) If the bird turns effectually and flies with the wind, how long volition he take to return half dozen.00 km? (c) Discuss how the air current affects the full round-trip time compared to what it would be with no air current.

3. Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of three.50 m/south, and the second a velocity of 4.xx thou/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 one thousand from the finish line, who will win the race, bold they run at constant velocity? (c) What distance ahead volition the winner be when she crosses the finish line?

4. Verify that the coin dropped past the airline passenger in Figure 7 travels 144 yard horizontally while falling one.50 m in the frame of reference of the Earth.

5. A football quarterback is moving straight astern at a speed of 2.00 m/southward when he throws a pass to a player xviii.0 one thousand straight downfield. The ball is thrown at an bending of 25.0º relative to the ground and is defenseless at the same elevation as information technology is released. What is the initial velocity of the brawl relative to the quarterback ?

6. A ship sets sail from Rotterdam, The Netherlands, heading due due north at seven.00 m/s relative to the water. The local bounding main current is one.50 k/due south in a management 40.0º north of east. What is the velocity of the ship relative to the Earth?

7. (a) A jet aeroplane flying from Darwin, Australia, has an air speed of 260 1000/s in a direction 5.0º south of west. It is in the jet stream, which is bravado at 35.0 m/s in a direction 15º south of east. What is the velocity of the airplane relative to the World? (b) Hash out whether your answers are consistent with your expectations for the effect of the wind on the plane'south path.

8. (a) In what direction would the ship in Exercise half-dozen accept to travel in order to have a velocity straight north relative to the World, bold its speed relative to the h2o remains seven.00 thousand/s? (b) What would its speed exist relative to the Earth?

9. (a) Another airplane is flight in a jet stream that is blowing at 45.0 g/s in a direction 20º s of east (as in Practise seven). Its direction of motion relative to the Globe is 45º south of west, while its direction of travel relative to the air is five.00º south of west. What is the airplane's speed relative to the air mass? (b) What is the aeroplane's speed relative to the Earth?

10. A sandal is dropped from the top of a xv.0-1000-high mast on a ship moving at i.75 k/south due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent event for the position at which the sandal hits the deck.

11. The velocity of the wind relative to the h2o is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 grand/s in a direction xxx.0º east of north relative to the Globe. It encounters a wind that has a velocity of 4.50 m/due south in a direction of 50.0º south of west relative to the World. What is the velocity of the wind relative to the water?

12. The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Style Galaxy with velocities proportional to their distances. It appears to an observer on the Globe that nosotros are at the heart of an expanding universe. Figure 9 illustrates this for five galaxies lying along a straight line, with the Milky Manner Milky way at the center. Using the data from the effigy, calculate the velocities: (a) relative to galaxy two and (b) relative to galaxy 5. The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely exist aware of relative velocities, final that it is non possible to locate the center of expansion with the given information.

Figure ix. Five galaxies on a straight line, showing their distances and velocities relative to the Milky Style (MW) Milky way. The distances are in millions of low-cal years (Mly), where a light yr is the altitude light travels in ane year. The velocities are nearly proportional to the distances. The sizes of the galaxies are profoundly exaggerated; an average milky way is most 0.ane Mly beyond.

(a) Use the distance and velocity data in [link] to observe the charge per unit of expansion as a function of distance.

(b) If you extrapolate back in fourth dimension, how long ago would all of the galaxies have been at approximately the aforementioned position? The two parts of this problem give you some idea of how the Hubble abiding for universal expansion and the time dorsum to the Big Blindside are determined, respectively.

13. An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 g/s relative to the water. He reaches the reverse side at a altitude xl k downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at remainder on the ground?

fourteen. A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is iv.80 m/s 5.00º west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the Us.)

15. An water ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the thespian is 29.0 m/s. The line between the center of the goal and the player makes a ninety.0º bending relative to his path as shown in Effigy 10. What angle must the puck's velocity make relative to the player (in his frame of reference) to hit the eye of the goal?

Effigy 10. An ice hockey player moving beyond the rink must shoot backward to give the puck a velocity toward the goal.

16. Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km in a higher place the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is at that place a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum tiptop? (d) Is the premise unreasonable or is the bachelor equation inapplicable? Explicate your answer.

17. Unreasonable Results A commercial airplane has an air speed of 280 m/southward due east and flies with a stiff tailwind. Information technology travels 3000 km in a direction 5º south of east in one.50 h. (a) What was the velocity of the aeroplane relative to the footing? (b) Calculate the magnitude and direction of the tailwind's velocity. (c) What is unreasonable most both of these velocities? (d) Which premise is unreasonable?

18. Construct Your Own Trouble Consider an airplane headed for a track in a cantankerous air current. Construct a problem in which yous calculate the bending the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the management of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Hash out whatsoever concluding minute maneuvers the airplane pilot might accept to perform in order for the plane to land with its wheels pointing straight down the runway.

Glossary

classical relativity:

the study of relative velocities in situations where speeds are less than about 1% of the speed of light—that is, less than 3000 km/s

relative velocity:

the velocity of an object as observed from a particular reference frame

relativity:

the study of how different observers moving relative to each other measure the same phenomenon

velocity:

speed in a given direction

vector addition:

the rules that utilise to calculation vectors together

Selected Solutions to Issues & Exercises

1. (a) 35.8 km, 45º south of east (b) 5.53 m/southward, 45º south of east (c) 56.i km, 45º south of east

iii. (a) 0.70 yard/s faster (b) Second runner wins (c) 4.17 m

5. 17.0 m/s, 22.1º

7.  (a) 230 m/s, viii.0º south of due west (b) The wind should make the plane travel slower and more to the south, which is what was calculated.

9. (a) 63.5 g/s (b) 29.half dozen m/south

11. six.68 m/southward, 53.3º south of w

12.  (a) [latex]{H}_{\text{average}}=\text{fourteen}\text{.}\text{9}\frac{\text{km/due south}}{\text{Mly}}[/latex] (b) twenty.2 billion years

xiv. 1.72 chiliad/s, 42.3º north of e

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Source: https://courses.lumenlearning.com/physics/chapter/3-5-addition-of-velocities/

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