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Inside Collection (Textbook):

Textbook by: Oka Kurniawan. E-mail the author

Vector Addition and Subtraction: Analytical Methods

Module by: OpenStax College. E-mail the author

Summary:

• Understand the rules of vector addition and subtraction using analytical methods.
• Apply analytical methods to determine vertical and horizontal component vectors.
• Apply analytical methods to determine the magnitude and direction of a resultant vector.

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like AA size 12{A} {} in Figure 1, we may wish to find which two perpendicular vectors, AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {}, add to produce it.

AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {} are defined to be the components of AA size 12{A} {} along the x- and y-axes. The three vectors AA size 12{A} {}, AxAx size 12{A rSub { size 8{x} } } {}, and AyAy size 12{A rSub { size 8{y} } } {} form a right triangle:

Ax + Ay = A.Ax + Ay = A. size 12{A rSub { size 8{x} } bold " + A" rSub { size 8{y} } bold " = A."} {}
(1)

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if Ax=3 mAx=3 m size 12{A rSub { size 8{x} } } {} east, Ay=4 mAy=4 m size 12{A rSub { size 8{y} } } {} north, and A=5 mA=5 m size 12{A} {} north-east, then it is true that the vectors Ax + Ay = AAx + Ay = A size 12{A rSub { size 8{x} } bold " + A" rSub { size 8{y} } bold " = A"} {}. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

3 m+4 m   5 m 3 m+4 m   5 m alignl { stack { size 12{"3 M + 4 M " <> " 5 M"} {} # {} } } {}
(2)

Thus,

A x + A y A A x + A y A size 12{A rSub { size 8{x} } +A rSub { size 8{y} } <> A} {}
(3)

If the vector AA size 12{A} {} is known, then its magnitude AA size 12{A} {} (its length) and its angle θ θ size 12{θ} {} (its direction) are known. To find AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {}, its x- and y-components, we use the following relationships for a right triangle.

A x = A cos θ A x = A cos θ size 12{A rSub { size 8{x} } =A"cos"θ} {}
(4)

and

Ay=Asinθ.Ay=Asinθ. size 12{A rSub { size 8{y} } =A"sin"θ"."} {}
(5)

Suppose, for example, that AA size 12{A} {} is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Then A=10.3A=10.3 size 12{A} {} blocks and θ = 29.1º θ = 29.1º size 12{"29.1º"} , so that

Ax=Acosθ=(10.3 blocks)(cos29.1º)=9.0 blocksAx=Acosθ=(10.3 blocks)(cos29.1º)=9.0 blocks size 12{}
(6)
Ay=Asinθ=(10.3 blocks)(sin29.1º)=5.0 blocks.Ay=Asinθ=(10.3 blocks)(sin29.1º)=5.0 blocks. size 12{""}
(7)

Calculating a Resultant Vector

If the perpendicular components AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {} of a vector AA size 12{A} {} are known, then AA size 12{A} {} can also be found analytically. To find the magnitude AA size 12{A} {} and direction θ θ size 12{θ} {} of a vector from its perpendicular components AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {}, we use the following relationships:

A=Ax2+Ay2A=Ax2+Ay2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {}
(8)
θ = tan 1 ( A y / A x ) . θ = tan 1 ( A y / A x ) . size 12{θ="tan" rSup { size 8{ - 1} } $$A rSub { size 8{y} } /A rSub { size 8{x} }$$ } {}
(9)

Note that the equation A = A x 2 + A y 2 A = A x 2 + A y 2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {} is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {} are 9 and 5 blocks, respectively, then A=92+52=10.3A=92+52=10.3 size 12{A= sqrt {9 rSup { size 8{2} } "+5" rSup { size 8{2} } } "=10" "." 3} {} blocks, again consistent with the example of the person walking in a city. Finally, the direction is θ = tan –1 ( 5/9 ) =29.1º θ = tan –1 ( 5/9 ) =29.1º size 12{θ="tan" rSup { size 8{–1} } $$"5/9"$$ "=29" "." 1 rSup { size 8{o} } } {} , as before.

Determining Vectors and Vector Components with Analytical Methods:

Equations Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {} are used to find the perpendicular components of a vector—that is, to go from AA size 12{A} {} and θ θ size 12{θ} {} to AxAx size 12{A rSub { size 8{x} } } {} and AyAy size 12{A rSub { size 8{y} } } {}. Equations A=Ax2+Ay2A=Ax2+Ay2 size 12{A= sqrt {A rSub { size 8{x} rSup { size 8{2} } } +A rSub { size 8{y} rSup { size 8{2} } } } } {} and θ=tan–1(Ay/Ax)θ=tan–1(Ay/Ax) are used to find a vector from its perpendicular components—that is, to go from AxAx and AyAy to AA and θ θ . Both processes are crucial to analytical methods of vector addition and subtraction.

To see how to add vectors using perpendicular components, consider Figure 5, in which the vectors AA size 12{A} {} and BB size 12{B} {} are added to produce the resultant RR size 12{R} {}.

If AA and BB represent two legs of a walk (two displacements), then RR is the total displacement. The person taking the walk ends up at the tip of R.R. There are many ways to arrive at the same point. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, RxRx and RyRy size 12{R rSub { size 8{y} } } {}. If we know RxRx and RyRy size 12{R rSub { size 8{y} } } {}, we can find RR and θ θ using the equations A = A x 2 + Ay 2 A = A x 2 + Ay 2 and θ =tan –1 (Ay /Ax )θ =tan –1 (Ay /Ax ) size 12{θ="tan" rSup { size 8{–1} } $$A rSub { size 8{y} } /A rSub { size 8{x} }$$ } {}. When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {} and Ay=AsinθAy=Asinθ size 12{A rSub { size 8{y} } =A"sin"θ} {} to find the components. In Figure 6, these components are AxAx size 12{A rSub { size 8{x} } } {}, AyAy size 12{A rSub { size 8{y} } } {}, BxBx size 12{B rSub { size 8{x} } } {}, and ByBy size 12{B rSub { size 8{y} } } {}. The angles that vectors AA size 12{A} {} and BB size 12{B} {} make with the x-axis are θAθA size 12{θ rSub { size 8{A} } } {} and θBθB size 12{θ rSub { size 8{B} } } {}, respectively.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 7,

R x = A x + B x R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}
(10)

and

R y = A y + B y . R y = A y + B y . size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}
(11)

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of RR size 12{R} {} are known, its magnitude and direction can be found.

Step 3. To get the magnitude RR size 12{R } {} of the resultant, use the Pythagorean theorem:

R=Rx2+Ry2.R=Rx2+Ry2. size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } "."} {}
(12)

Step 4. To get the direction of the resultant:

θ=tan1(Ry/Rx).θ=tan1(Ry/Rx). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ "."} {}
(13)

The following example illustrates this technique for adding vectors using perpendicular components.

Example 1: Adding Vectors Using Analytical Methods

Add the vector AA size 12{A} {} to the vector BB size 12{B} {} shown in Figure 8, using perpendicular components along the x- and y-axes. The x- and y-axes are along the east–west and north–south directions, respectively. Vector AA size 12{A} {} represents the first leg of a walk in which a person walks 53.0 m53.0 m size 12{"53" "." "0 m"} {} in a direction 20.0º20.0º size 12{"20" "." 0º } {} north of east. Vector BB size 12{B} {} represents the second leg, a displacement of 34.0 m34.0 m size 12{"34" "." "0 m"} {} in a direction 63.0º63.0º size 12{"63" "." 0º } {} north of east.

Strategy

The components of AA size 12{A} {} and BB size 12{B} {} along the x- and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of AA size 12{A} {} and BB size 12{B} {} along the x- and y-axes. Note that A=53.0 mA=53.0 m size 12{"A" "=" "53.0 m"} {}, θA=20.0ºθA=20.0º size 12{"θ" "subA" "=" "20.0°" } {}, B=34.0 mB=34.0 m size 12{"B" "=" "34.0" "m"} {}, and θB=63.0ºθB=63.0º size 12{θ rSub { size 8{B} } } {}. We find the x-components by using Ax=AcosθAx=Acosθ size 12{A rSub { size 8{x} } =A"cos"θ} {}, which gives

A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ rSub { size 8{A} } = $$"53" "." 0" m"$$ $$"cos""20" "." 0 { size 12{ circ } }$$ } {} # " "= $$"53" "." 0" m"$$ $$0 "." "940"$$ ="49" "." 8" m" {} } } {}
(14)

and

B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . alignl { stack { size 12{B rSub { size 8{x} } =B"cos"θ rSub { size 8{B} } = $$"34" "." 0" m"$$ $$"cos""63" "." 0 { size 12{ circ } }$$ } {} # " "= $$"34" "." 0" m"$$ $$0 "." "454"$$ ="15" "." 4" m" {} } } {}
(15)

Similarly, the y-components are found using Ay=AsinθAAy=AsinθA size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } } {}:

A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m alignl { stack { size 12{A rSub { size 8{y} } =A"sin"θ rSub { size 8{A} } = $$"53" "." 0" m"$$ $$"sin""20" "." 0 { size 12{ circ } }$$ } {} # " "= $$"53" "." 0" m"$$ $$0 "." "342"$$ ="18" "." 1" m" {} } } {}
(16)

and

B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . alignl { stack { size 12{B rSub { size 8{y} } =B"sin"θ rSub { size 8{B} } = $$"34" "." 0" m"$$ $$"sin""63" "." 0 { size 12{ circ } }$$ } {} # " "= $$"34" "." 0" m"$$ $$0 "." "891"$$ ="30" "." 3" m" "." {} } } {}
(17)

The x- and y-components of the resultant are thus

R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } ="49" "." 8" m"+"15" "." 4" m"="65" "." 2" m"} {}
(18)

and

Ry=Ay+By=18.1 m+30.3 m=48.4 m.Ry=Ay+By=18.1 m+30.3 m=48.4 m. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } ="18" "." 1" m"+"30" "." 3" m"="48" "." 4" m."} {}
(19)

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } = sqrt { $$"65" "." 2$$ rSup { size 8{2} } + $$"48" "." 4$$ rSup { size 8{2} } } " m"} {}
(20)

so that

R = 81.2 m. R = 81.2 m. size 12{R ="81.2" "m."} {}
(21)

Finally, we find the direction of the resultant:

θ=tan1(Ry/Rx)=+tan1(48.4/65.2).θ=tan1(Ry/Rx)=+tan1(48.4/65.2). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ "=+""tan" rSup { size 8{ - 1} } $$"48" "." 4/"65" "." 2$$ "."} {}
(22)

Thus,

θ=tan1(0.742)=36.6º.θ=tan1(0.742)=36.6º. size 12{θ="tan" rSup { size 8{ - 1} } $$0 "." "742"$$ ="36" "." 6 { size 12{ circ } } "."} {}
(23)

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, ABA+(–B)ABA+(–B) size 12{A – B equiv A+ $$- B$$ } {}. Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of –B–B are the negatives of the components of BB size 12{B} {}. The x- and y-components of the resultant AB = RAB = R size 12{A- bold "B = R"} {} are thus

R x = A x + ( B x ) R x = A x + ( B x ) size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +-B rSub { size 8{x} } } {}
(24)

and

R y = A y + ( B y ) R y = A y + ( B y ) size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +-B rSub { size 8{y} } } {}
(25)

and the rest of the method outlined above is identical to that for addition. (See Figure 10.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

Summary

• The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
• The steps to add vectors AA size 12{A} {} and BB size 12{B} {} using the analytical method are as follows:

Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

A x = A cos θ B x = B cos θ A x = A cos θ B x = B cos θ alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ} {} # B rSub { size 8{x} } =B"cos"θ {} } } {}
(26)

and

A y = A sin θ By = B sin θ . A y = A sin θ By = B sin θ . alignl { stack { size 12{A rSub { size 8{y} } =A" sin"θ} {} # B=B suby " sin "θ {} } } {}
(27)

Step 2: Add the horizontal and vertical components of each vector to determine the components RxRx size 12{R rSub { size 8{x} } } {} and RyRy size 12{R rSub { size 8{y} } } {} of the resultant vector, RR size 12{R} {}:

R x = A x + B x R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}
(28)

and

Ry=Ay+By.Ry=Ay+By. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}
(29)

Step 3: Use the Pythagorean theorem to determine the magnitude, RR size 12{R} {}, of the resultant vector RR size 12{R} {}:

R=Rx2+Ry2.R=Rx2+Ry2. size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } } {}
(30)

Step 4: Use a trigonometric identity to determine the direction, θθ size 12{θ} {}, of RR size 12{R} {}:

θ=tan1(Ry/Rx).θ=tan1(Ry/Rx). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ } {}
(31)

Conceptual Questions

Exercise 1

Suppose you add two vectors AA size 12{A} {} and BB size 12{B} {}. What relative direction between them produces the resultant with the greatest magnitude? What is the maximum magnitude? What relative direction between them produces the resultant with the smallest magnitude? What is the minimum magnitude?

Exercise 2

Give an example of a nonzero vector that has a component of zero.

Exercise 3

Explain why a vector cannot have a component greater than its own magnitude.

Exercise 4

If the vectors AA size 12{A} {} and BB size 12{B} {} are perpendicular, what is the component of AA size 12{A} {} along the direction of BB size 12{B} {}? What is the component of BB size 12{B} {} along the direction of AA size 12{A} {}?

Problems & Exercises

Exercise 1

Find the following for path C in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

(a) 1.56 km

(b) 120 m east

Exercise 2

Find the following for path D in Figure 12: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

Exercise 3

Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 13.

Solution

North-component 87.0 km, east-component 87.0 km

Exercise 4

Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements AA size 12{A} {} and BB size 12{B} {}, as in Figure 14, then this problem asks you to find their sum R=A+BR=A+B size 12{R=A+B} {}.)

Note that you can also solve this graphically. Discuss why the analytical technique for solving this problem is potentially more accurate than the graphical technique.

Exercise 5

Repeat Exercise 4 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking you other path.

Solution

30.8 m, 35.8 west of north

Exercise 6

You drive 7.50 km7.50 km size 12{7 "." "50 km"} {} in a straight line in a direction 15º15º size 12{"15º"} {} east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

Exercise 7

Do Exercise 4 again using analytical techniques and change the second leg of the walk to 25.0 m25.0 m straight south. (This is equivalent to subtracting BB size 12{B} {} from AA size 12{A} {} —that is, finding R=A – BR=A – B) (b) Repeat again, but now you first walk 25.0 m25.0 m size 12{"25" "." "0 m"} {} north and then 18.0 m18.0 m size 12{"18" "." "0 m"} {} east. (This is equivalent to subtract AA size 12{A} {} from BB size 12{B} {} —that is, to find A=B+CA=B+C size 12{A=B+C} {}. Is that consistent with your result?)

Solution

(a) 30.8 m30.8 m size 12{"30" "." "8 m"} {}, 54.54. size 12{"54" "." 2°} {} south of west

(b) 30.8 m30.8 m size 12{"30" "." "8 m"} {}, 54.54. size 12{"54" "." 2°} {} north of east

Exercise 8

A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors AA size 12{A} {} from BB size 12{B} {} in Figure 15. She then correctly calculates the length and orientation of the third side CC size 12{C} {}. What is her result?

Exercise 9

You fly 32.0 km32.0 km size 12{"32" "." "0 km"} {} in a straight line in still air in the direction 35.0º35.0º size 12{"35"°} {} south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º45.0º size 12{"45.0º"} {} south of west and then in a direction 45.0º45.0º size 12{"45.0º"} {} west of north. These are the components of the displacement along a different set of axes—one rotated 45º45º size 12{"45"°} {}.

Solution

18.4 km south, then 26.2 km west(b) 31.5 km at 45.0º45.0º size 12{"45.0º"} {} south of west, then 5.56 km at 45.0º45.0º size 12{"45.0º"} {} west of north

Exercise 10

A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A , A , size 12{A,} {} B , B , size 12{B,} {} and C C size 12{C} {} in Figure 16, and then correctly calculates the length and orientation of the fourth side DD size 12{D} {}. What is his result?

Exercise 11

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km2.50 km size 12{2 "." "50 km"} {} 45.0º45.0º size 12{"45.0º"} {} north of west; then 4.70 km4.70 km size 12{4 "." "70 km"} {} 60.0º60.0º size 12{"60"°} {} south of east; then 1.30 km1.30 km size 12{"1.30" " km"} {} 25.0º25.0º size 12{"25"°} {} south of west; then 5.10 km5.10 km size 12{5 "." "10 km"} {} straight east; then 1.70 km1.70 km size 12{"1.70" " km"} {} 5.00º5.00º size 12{5 rSup { size 8{ circ } } } {} east of north; then 7.20 km 7.20 km size 12{7 "." "20 km"} {} 55.0º55.0º size 12{"55"°} {} south of west; and finally 2.80 km2.80 km size 12{2 "." "80 km"} {} 10.0º10.0º size 12{"10"°} {} north of east. What is his final position relative to the island?

Solution

7.34 km7.34 km size 12{2 "." "97 km"} {}, 63.63. size 12{"22" "." 2°} {} south of east

Exercise 12

Suppose a pilot flies 40.0 km40.0 km size 12{"40" "." "0 km"} {} in a direction 60º60º size 12{"60"°} {} north of east and then flies 30.0 km30.0 km size 12{"30" "." "0 km"} {} in a direction 15º15º size 12{"15"°} {} north of east as shown in Figure 17. Find her total distance RR size 12{R} {} from the starting point and the direction θθ size 12{θ} {} of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

Glossary

analytical method:
the method of determining the magnitude and direction of a resultant vector using the Pythagorean theorem and trigonometric identities

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