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Kinematics

18 Vector Addition and Subtraction: Graphical Methods

Learning Objectives

  • Understand the rules of vector addition, subtraction, and multiplication.

  • Apply graphical methods of vector addition and subtraction to determine the displacement of moving objects.

 

Some Hawaiian Islands like Kauai Oahu, Molokai, Lanai, Maui, Kahoolawe, and Hawaii are shown. On the scale map of Hawaiian Islands the path of a journey is shown moving from Hawaii to Molokai. The path of the journey is turning at different angles and finally reaching its destination. The displacement of the journey is shown with the help of a straight line connecting its starting point and the destination.
Figure 18.1: Displacement determined graphically by adding journey segments on a scale map of the Hawaiian Islands. (credit: US Geological Survey)

Vectors in Two Dimensions

A vector is a quantity with both magnitude and direction. Examples include displacement, velocity, acceleration, and force. In one-dimensional motion, direction can be expressed with a plus or minus sign. In two dimensions, direction is specified relative to a reference frame or coordinate system by an arrow whose length is proportional to magnitude and points in the vector’s direction.

In this text, vectors are represented by boldface symbols, for example, the displacement vector is D. Its magnitude is written in italics as D, and its direction is given by an angle [latex]\theta[/latex].

 

A graph is shown. On the axes the scale is set to one block is equal to one unit. A helicopter starts moving from the origin at an angle of twenty nine point one degrees above the x axis. The current position of the helicopter is ten point three blocks along its line of motion. The destination of the helicopter is the point which is nine blocks in the positive x direction and five blocks in the positive y direction. The positive direction of the x axis is east and the positive direction of the y axis is north.
Figure 18.2: Displacement vector for a person walking 9 blocks east and 5 blocks north: magnitude D = 10.3 blocks, direction [latex]\theta = 29.1º[/latex] north of east.
On a graph a vector is shown. It is inclined at an angle theta equal to twenty nine point one degrees above the positive x axis. A protractor is shown to the right of the x axis to measure the angle. A ruler is also shown parallel to the vector to measure its length. The ruler shows that the length of the vector is ten point three units.
Figure 18.3: Resultant displacement vector D drawn at angle [latex]\theta[/latex] relative to east-west axis, with magnitude proportional to length.

Vector Addition: Head-to-Tail Method

The head-to-tail method described in Figure 18.4  is a graphical way to add vectors:

  1. Draw the first vector with arrow length proportional to magnitude.

  2. From the head (tip) of the first vector, draw the second vector.

  3. For multiple vectors, continue placing each new vector’s tail at the previous vector’s head.

  4. Draw the resultant vector from the tail of the first vector to the head of the last vector.

  5. Measure the resultant’s length for magnitude.

  6. Measure the angle it makes with a reference axis for direction.

In part a, a vector of magnitude of nine units and making an angle of theta is equal to zero degrees is drawn from the origin and along the positive direction of x axis. In part b a vector of magnitude of nine units and making an angle of theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical arrow from the head of the horizontal arrow is drawn. In part c a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vectorD.
Figure 18.4: Adding two displacement vectors head-to-tail and drawing the resultant vector D.

Step 1.Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.

In part a, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis.
Figure 18.5

Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector.

In part b, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical vector from the head of the horizontal vector is drawn.
Figure 18.6

Step 3.If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail.

Step 4.Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors.

In part c, a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of the x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vector D.
Figure 18.7

Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.

Example 18.1: Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk

Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction [latex]\text{49.0º}[/latex] north of east. Then, she walks 23.0 m heading [latex]\text{15.0º}[/latex] north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.

Strategy

Represent each displacement vector graphically with an arrow, labeling the first [latex]\text{A}[/latex], the second [latex]\text{B}[/latex], and the third [latex]\text{C}[/latex], making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted [latex]\mathbf{\text{R}}[/latex].

Solution

  1. Draw the three displacement vectors.
    On the graph a vector of magnitude twenty three meters and inclined above the x axis at an angle theta-b equal to fifteen degrees is shown. This vector is labeled as B.
    Figure 18.8
  2. Place the vectors head to tail retaining both their initial magnitude and direction.
    In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis.
    Figure 18.9
  3. Draw the resultant vector, [latex]\text{R}[/latex].
    In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis. From the tail of the vector A a vector R of magnitude of fifty point zero meters and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector C. The vector R is known as the resultant vector.
    Figure 18.10
  4. Use a ruler to measure the magnitude of [latex]\mathbf{\text{R}}[/latex], and a protractor to measure the direction of [latex]\text{R}[/latex]. While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.
    In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis. From the tail of the vector A a vector R of magnitude of fifty meter and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector C. The vector R is known as the resultant vector. A ruler is placed along the vector R to measure it. Also there is a protractor to measure the angle.
    Figure 18.11

    In this case, the total displacement [latex]\mathbf{\text{R}}[/latex] is seen to have a magnitude of 50.0 m and to lie in a direction [latex]7.0º[/latex] south of east. By using its magnitude and direction, this vector can be expressed as [latex]R=\text{50.0 m}[/latex] and [latex]\theta =7\text{.}\text{0º}[/latex] south of east.

Discussion

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 18.12 and we will still get the same solution.

In this figure a vector C with a negative slope is drawn from the origin. Then from the head of the vector C another vector A with positive slope is drawn and then another vector B with negative slope from the head of the vector A is drawn. From the tail of the vector C a vector R of magnitude of fifty point zero meters and with negative slope of seven degrees is drawn. The head of this vector R meets the head of the vector B. The vector R is known as the resultant vector.
Figure 18.12

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.

[latex]\mathbf{\text{A}}+\mathbf{\text{B}}=\mathbf{\text{B}}+\mathbf{\text{A}}\text{.}[/latex]

(This is true for the addition of ordinary numbers as well—you get the same result whether you add [latex]\mathbf{\text{2}}+\mathbf{\text{3}}[/latex] or [latex]\mathbf{\text{3}}+\mathbf{\text{2}}[/latex], for example).

Vector Subtraction

Vector subtraction is adding the negative of a vector:

  • The negative of vector B is vector –B, which has the same magnitude but points in the opposite direction.

  • Subtraction defined as:
    [latex]\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})[/latex]

The negative of a vector [latex]\mathbf{\text{B}}[/latex] is defined to be [latex]\mathbf{\text{–B}}[/latex]; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 18.13. In other words, [latex]\mathbf{\text{B}}[/latex] has the same length as [latex]\mathbf{\text{–B}}[/latex], but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

Two vectors are shown. One of the vectors is labeled as vector in north east direction. The other vector is of the same magnitude and is in the opposite direction to that of vector B. This vector is denoted as negative B.
Figure 18.13: The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So [latex]\mathbf{\text{B}}[/latex] is the negative of [latex]\mathbf{\text{–B}}[/latex]; it has the same length but opposite direction.

This is analogous to the subtraction of scalars (where, for example, [latex]\text{5 – 2 = 5 + }\left(\text{–2}\right)[/latex]). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Example 18.2: Subtracting Vectors Graphically: A Woman Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction [latex]\text{66.0º}[/latex] north of east from her current location, and then travel 30.0 m in a direction [latex]\text{112º}[/latex] north of east (or [latex]\text{22.0º}[/latex] west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.

A vector of magnitude twenty seven point five meters is shown. It is inclined to the horizontal at an angle of sixty six degrees. Another vector of magnitude thirty point zero meters is shown. It is inclined to the horizontal at an angle of one hundred and twelve degrees.
Figure 18.14

Strategy

We can represent the first leg of the trip with a vector [latex]\mathbf{\text{A}}[/latex], and the second leg of the trip with a vector [latex]\mathbf{\text{B}}[/latex]. The dock is located at a location [latex]\mathbf{\text{A}}+\mathbf{\text{B}}[/latex]. If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance [latex]B[/latex] (30.0 m) in the direction [latex]180º–112º=68º[/latex] south of east. We represent this as [latex]\mathbf{\text{–B}}[/latex], as shown below. The vector [latex]\mathbf{\text{–B}}[/latex] has the same magnitude as [latex]\mathbf{\text{B}}[/latex] but is in the opposite direction. Thus, she will end up at a location [latex]\mathbf{\text{A}}+\left(\mathbf{\text{–B}}\right)[/latex], or [latex]\mathbf{\text{A}}–\mathbf{\text{B}}[/latex].

A vector labeled negative B is inclined at an angle of sixty-eight degrees below a horizontal line. A dotted line in the reverse direction inclined at one hundred and twelve degrees above the horizontal line is also shown.
Figure 18.15

We will perform vector addition to compare the location of the dock, [latex]\text{A }\text{+ }\mathbf{B}[/latex], with the location at which the woman mistakenly arrives, [latex]\text{A + }\left(\text{–B}\right)[/latex].

Solution

  1. To determine the location at which the woman arrives by accident, draw vectors [latex]\mathbf{\text{A}}[/latex] and [latex]\mathbf{\text{–B}}[/latex].
  2. Place the vectors head to tail.
  3. Draw the resultant vector [latex]\mathbf{R}[/latex].
  4. Use a ruler and protractor to measure the magnitude and direction of [latex]\mathbf{R}[/latex].
    Vectors A and negative B are connected in head to tail method. Vector A is inclined with horizontal with positive slope and vector negative B with a negative slope. The resultant of these two vectors is shown as a vector R from tail of A to the head of negative B. The length of the resultant is twenty three point zero meters and has a negative slope of seven point five degrees.
    Figure 18.16

    In this case, [latex]R=\text{23}\text{.}\text{0 m}[/latex] and [latex]\theta =7\text{.}\text{5º}[/latex] south of east.

  5. To determine the location of the dock, we repeat this method to add vectors [latex]\mathbf{\text{A}}[/latex] and [latex]\mathbf{\text{B}}[/latex]. We obtain the resultant vector [latex]\mathbf{\text{R}}\text{'}[/latex]:
    A vector A inclined at sixty six degrees with horizontal is shown. From the head of this vector another vector B is started. Vector B is inclined at one hundred and twelve degrees with the horizontal. Another vector labeled as R prime from the tail of vector A to the head of vector B is drawn. The length of this vector is fifty two point nine meters and its inclination with the horizontal is shown as ninety point one degrees. Vector R prime is equal to the sum of vectors A and B.
    Figure 18.17

    In this case [latex]R\text{ = 52.9 m}[/latex] and [latex]\theta =\text{90.1º}[/latex] north of east.

    We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Multiplication of Vectors and Scalars

  • Multiplying a vector by a positive scalar changes magnitude but not direction.

  • Multiplying by a negative scalar reverses the vector’s direction and changes magnitude.

For vector A multiplied by scalar c:

  • Magnitude becomes [latex]|c| A[/latex]

  • Direction unchanged if [latex]c > 0[/latex]

  • Direction reversed if [latex]c 0[/latex]

Example: Walking three times the distance of 27.5 m at 66.0º north of east results in 82.5 m in the same direction.

Resolving a Vector into Components

Sometimes we want to find the perpendicular components (e.g., east and north) of a single vector.

Example: A total displacement of 10.3 blocks at 29.0º north of east can be broken down into:

  • Eastward component = ?

  • Northward component = ?

This method is essential in physics (e.g., projectile motion, forces) and involves right triangles and trigonometry.

PhET Explorations: Maze Game

Learn about position, velocity, and acceleration in the “Arena of Pain”. Use the green arrow to move the ball. Add more walls to the arena to make the game more difficult. Try to make a goal as fast as you can.

Figure 18.18: Maze Game

Summary

  • The graphical method of adding vectors [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex] involves drawing the vectors on a graph and using the head-to-tail method. The resultant vector [latex]\mathbf{R}[/latex] is defined by

    [latex]\mathbf{A} + \mathbf{B} = \mathbf{R}.[/latex]

    The magnitude and direction of [latex]\mathbf{R}[/latex] are then measured using a ruler and protractor, respectively.

  • The graphical method of subtracting vector [latex]\mathbf{B}[/latex] from [latex]\mathbf{A}[/latex] involves adding the negative of [latex]\mathbf{B}[/latex], defined as [latex]-\mathbf{B}[/latex]. Thus,

    [latex]\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B}) = \mathbf{R}.[/latex]

    Then, the usual head-to-tail method is used to find the resultant vector [latex]\mathbf{R}[/latex].

  • Vector addition is commutative:

    [latex]\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}.[/latex]

  • The head-to-tail method involves drawing the first vector on a graph, then placing the tail of each subsequent vector at the head of the previous one. The resultant vector is drawn from the tail of the first vector to the head of the last vector.

  • If a vector [latex]\mathbf{A}[/latex] is multiplied by a scalar [latex]c[/latex], the magnitude of the product is

    [latex]|c| A.[/latex]

    If [latex]c[/latex] is positive, the product points in the same direction as [latex]\mathbf{A}[/latex]; if [latex]c[/latex] is negative, it points in the opposite direction.

Conceptual Questions

  1. Which of the following is a vector: a person’s height, the altitude on Mt. Everest, the age of the Earth, the boiling point of water, the cost of this book, the Earth’s population, the acceleration of gravity?
  2. Give a specific example of a vector, stating its magnitude, units, and direction.
  3. What do vectors and scalars have in common? How do they differ?
  4. Two campers in a national park hike from their cabin to the same spot on a lake, each taking a different path, as illustrated below. The total distance traveled along Path 1 is 7.5 km, and that along Path 2 is 8.2 km. What is the final displacement of each camper?
    At the southwest corner of the figure is a cabin and in the northeast corner is a lake. A vector S with a length five point zero kilometers connects the cabin to the lake at an angle of 40 degrees north of east. Two winding paths labeled Path 1 and Path 2 represent the routes travelled from the cabin to the lake.
    Figure 18.19

  5. If an airplane pilot is told to fly 123 km in a straight line to get from San Francisco to Sacramento, explain why he could end up anywhere on the circle shown in Figure 18.20. What other information would he need to get to Sacramento?

     

    A map of northern California with a circle with a radius of one hundred twenty three kilometers centered on San Francisco. Sacramento lies on the circumference of this circle in a direction forty-five degrees north of east from San Francisco.
    Figure 18.20
  6. Suppose you take two steps [latex]\mathbf{\text{A}}[/latex] and [latex]\mathbf{\text{B}}[/latex] (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point [latex]\mathbf{\text{A}}+\mathbf{\text{B}}[/latex] the sum of the lengths of the two steps?
  7. Explain why it is not possible to add a scalar to a vector.
  8. If you take two steps of different sizes, can you end up at your starting point? More generally, can two vectors with different magnitudes ever add to zero? Can three or more?

Problems & Exercises

Use graphical methods to solve these problems. You may assume data taken from graphs is accurate to three digits.

  1. Find the following for path A in Figure 18.21:
    A map of city is shown. The houses are in form of square blocks of side one hundred and twenty meters each. The path of A extends to three blocks towards north and then one block towards east. It is asked to find out the total distance traveled the magnitude and the direction of the displacement from start to finish.
    Figure 18.21: The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.
    1. the total distance traveled
    2. the magnitude and direction of the displacement from start to finish.

      (a) [latex]\text{480 m}[/latex]
      (b) [latex]\text{379 m}[/latex], [latex]\text{18.4º}[/latex] east of north

  2. Find the following for path B in Figure 18.21
    1. (a) the total distance traveled
    2. the magnitude and direction of the displacement from start to finish.
  3. Find the north and east components of the displacement for the hikers shown in Figure 18.19
    north component 3.21 km, east component 3.83 km
  4. 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 [latex]\mathbf{\text{A}}[/latex] and [latex]\mathbf{\text{B}}[/latex], as in Figure 18.22, then this problem asks you to find their sum [latex]\mathbf{\text{R}}=\mathbf{\text{A}}+\mathbf{\text{B}}[/latex].)
    In this figure coordinate axes are shown. Vector A from the origin towards the negative of x axis is shown. From the head of the vector A another vector B is drawn towards the positive direction of y axis. The resultant R of these two vectors is shown as a vector from the tail of vector A to the head of vector B. This vector R is inclined at an angle theta with the negative x axis.
    Figure 18.22: The two displacements [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex] add to give a total displacement [latex]\mathbf{R}[/latex] having magnitude [latex]R[/latex] and direction [latex]\theta [/latex].
  5. Suppose you first walk 12.0 m in a direction [latex]\text{20º}[/latex] west of north and then 20.0 m in a direction [latex]\text{40.0º}[/latex] south of west. 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 [latex]\mathbf{A}[/latex] and [latex]\mathbf{B}[/latex], as in Figure 18.23, then this problem finds their sum [latex]\text{R = A + B}[/latex].)
    In the given figure coordinates axes are shown. Vector A with tail at origin is inclined at an angle of twenty degrees with the positive direction of x axis. The magnitude of vector A is twelve meters. Another vector B is starts from the head of vector A and inclined at an angle of forty degrees with the horizontal. The resultant R of the vectors A and B is also drawn from the tail of vector A to the head of vector B. The inclination of vector R is theta with the horizontal.
    Figure 18.23

    [latex]\text{19}\text{.}\text{5 m}[/latex], [latex]4\text{.}\text{65º}[/latex] south of west

  6. Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg [latex]\mathbf{B}[/latex], which is 20.0 m in a direction exactly [latex]\text{40º}[/latex] south of west, and then leg [latex]\mathbf{A}[/latex], which is 12.0 m in a direction exactly [latex]\text{20º}[/latex] west of north. (This problem shows that [latex]\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}[/latex].)
  7. Answer the following questions
    1. Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction [latex]\text{40.0º}[/latex] north of east (which is equivalent to subtracting [latex]\mathbf{\text{B}}[/latex] from [latex]\mathbf{A}[/latex] —that is, to finding [latex]\mathbf{\text{R}}\prime =\mathbf{\text{A}}-\mathbf{\text{B}}[/latex]).
    2. Repeat the problem two problems prior, but now you first walk 20.0 m in a direction [latex]\text{40.0º}[/latex] south of west and then 12.0 m in a direction [latex]\text{20.0º}[/latex] east of south (which is equivalent to subtracting [latex]\mathbf{\text{A}}[/latex] from [latex]\mathbf{\text{B}}[/latex] —that is, to finding [latex]\mathbf{\text{R}}\prime \prime =\mathbf{\text{B}}-\mathbf{\text{A}}=-\mathbf{\text{R}}\prime[/latex]). Show that this is the case.
      (a) [latex]\text{26}\text{.}\text{6 m}[/latex], [latex]\text{65}\text{.}\text{1º}[/latex] north of east
      (b) [latex]\text{26}\text{.}\text{6 m}[/latex], [latex]\text{65}\text{.}\text{1º}[/latex] south of west
  8. Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors [latex]\mathbf{A}[/latex], [latex]\mathbf{B}[/latex], and [latex]\mathbf{C}[/latex], all having different lengths and directions. Find the sum [latex]\text{A + B + C}[/latex] then find their sum when added in a different order and show the result is the same. (There are five other orders in which [latex]\mathbf{A}[/latex], [latex]\mathbf{B}[/latex], and [latex]\mathbf{C}[/latex] can be added; choose only one.)
  9. Show that the sum of the vectors discussed in Example 18.2 gives the result shown in Figure 18.17.
    [latex]\text{52}\text{.}\text{9 m}[/latex], [latex]\text{90}\text{.}\text{1º}[/latex] with respect to the x-axis.
  10. Find the magnitudes of velocities [latex]{v}_{\text{A}}[/latex] and [latex]{v}_{\text{B}}[/latex] in Figure 18.24.
    On the graph velocity vector V sub A begins at the origin and is inclined to x axis at an angle of twenty two point five degrees. From the head of vector V sub A another vector V sub B begins. The resultant of the two vectors, labeled V sub tot, is inclined to vector V sub A at twenty six point five degrees and to the vector V sub B at twenty three point zero degrees. V sub tot has a magnitude of 6.72 meters per second.
    Figure 18.24: The two velocities [latex]{\mathbf{\text{v}}}_{\text{A}}[/latex] and [latex]{\mathbf{\text{v}}}_{\text{B}}[/latex] add to give a total [latex]{\mathbf{\text{v}}}_{\text{tot}}[/latex].
  11. Find the components of [latex]{v}_{\text{tot}}[/latex] along the x– and y-axes in Figure 18.24.
    x-component 4.41 m/s
    y-component 5.07 m/s
  12. Find the components of [latex]{v}_{\text{tot}}[/latex] along a set of perpendicular axes rotated [latex]\text{30º}[/latex] counterclockwise relative to those in Figure 18.24.

Glossary

component (of a 2-d vector)
a piece of a vector that points in either the vertical or the horizontal direction; every 2-d vector can be expressed as a sum of two vertical and horizontal vector components
commutative
refers to the interchangeability of order in a function; vector addition is commutative because the order in which vectors are added together does not affect the final sum
direction (of a vector)
the orientation of a vector in spacecomponent (of a 2-d vector)
head (of a vector)
the end point of a vector; the location of the tip of the vector’s arrowhead; also referred to as the “tip”
head-to-tail method
a method of adding vectors in which the tail of each vector is placed at the head of the previous vector
magnitude (of a vector)
the length or size of a vector; magnitude is a scalar quantity
resultant
the sum of two or more vectors
resultant vector
the vector sum of two or more vectors
scalar
a quantity with magnitude but no direction
tail
the start point of a vector; opposite to the head or tip of the arrow
definition

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