1. Skipped to slide 40 for 2008 and 2009
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2. Introduction to Vectors
Scalars and Vectors
In Physics, quantities are described as either scalar quantities or vector quantities .
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3. Introduction to Vectors
Scalar Quantities
Involve only a magnitude, which includes numbers and units.
Examples include distance and speed.
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4. Introduction to Vectors
Vector Quantities
Involve a direction, in addition to numbers and units.
Can be represented graphically with arrows.
The longer the arrow, the greater the magnitude it represents.
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5. Introduction to Vectors
15 m/s east
25 m/s west
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6. Vector Addition in One direction
When vector quantities are in the same direction, vectors are added by placing the tail of one vector at the head of the other vector.
Be sure to maintain direction and length of vectors!
This creates one Resultant vector (R) which is drawn from the tail of the 1st vector to the head of the second vector.

7. Example: A child walks 2. 0 m east, pauses, and then continues 3
Example: A child walks 2.0 m east, pauses, and then continues 3.0 m east.
The resultant (R) = 5.0 m east.

8. If the two vectors have different directions, they are still added head to tail.
Example: A child walks 2.0 m east, then turns around and walks 4.0 m west.

9. Adding vectors graphically, using the “tip-to-tail” method.
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10. Component vectors are added “tip-to-tail.”
The resultant vector is drawn “tail-to-tip.”
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11. Adding vectors graphically, using the “tip-to-tail” method.
4 m east
3 m north
Adding vectors graphically, using the “tip-to-tail” method.
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12. A man walks 3 m north, and then 4 m east. Find his displacement.
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13. 4 m east
3 m north
You are allowed to move the vectors, but don’t change the direction or length.
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14. Line up the tip of one vector with the tail of the other.
4 m east
3 m north
Line up the tip of one vector with the tail of the other.
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15. Line up the tip of one vector with the tail of the other.
4 m east
3 m north
Line up the tip of one vector with the tail of the other.
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16. Now, draw the resultant vector from “tail-to-tip” as shown above.
4 m east
3 m north
Resultant vector
Now, draw the resultant vector from “tail-to-tip” as shown above.
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17. Remember to line up the component vectors from tip-to tail.
4 m east
3 m north
Remember to line up the component vectors from tip-to tail.
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18. If you line them up incorrectly, you get the wrong resultant vector.
4 m east
3 m north
If you line them up incorrectly, you get the wrong resultant vector.
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19. This is wrong! 4 m east How did this man go 3 m north
East, and then North,
and end up back where
he started!?!?
3 m north
Now, the resultant vector will be wrong, no matter how it is drawn.
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20. 4 m east
3 m north
Always line up the tip of one component vector with the tail of the other.
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21. Then draw your resultant vector from “tail-to-tip” as shown below.
4 m east
3 m north
Resultant vector
Then draw your resultant vector from “tail-to-tip” as shown below.
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22. It doesn’t matter in what order you add the component vectors.
4 m east
3 m north
It doesn’t matter in what order you add the component vectors.
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23. You will still get the same resultant vector.
3 m north
4 m east
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24. Compare two sets of identical component vectors.B B A
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25. It doesn’t matter which order you place the component vectors.B B A
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26. Only that you draw them from tip-to-tail.B B A
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27. The resultant vector will be the same in either case.
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28. The resultant vector will be the same in either case.
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29. Of course, we can add more than two component vectors.
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30. As long as we add them tip-to-tail.
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31. We will find the correct resultant vector.
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32. We can also subtract vectors graphically.
Find the resultant vector of vector A – vector B
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33. Instead of subtracting B from A, we add –B to A.
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34. A – B is the same as A + (-B)
Find the resultant vector A + (-B)
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35. -B B A
The vector called “-B” has the same magnitude as vector B, but the opposite direction.
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36. Now we add Vector A and Vector –B with the tip-to-tail method
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37. Now we add Vector A and Vector –B with the tip-to-tail method
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38. And draw the resultant vector.-B
Resultant vector
And draw the resultant vector.
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39. IF skip 3A beginning lecture (2008 and 2009)…
IF skip 3A beginning lecture (2008 and 2009)….use the next slides from lecture one and assign 3A
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40. Please complete the maze on your desk.
Happy Wednesday!!
Please complete the maze on your desk.
Both sides!
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41. Introduction to Vectors
Scalars and Vectors
In Physics, quantities are described as either scalar quantities or vector quantities .
12/31/2018 7:09 PM

42. Introduction to Vectors
Scalar Quantities
Involve only a magnitude, which includes numbers and units.
Examples include distance and speed.
12/31/2018 7:09 PM

43. Introduction to Vectors
Vector Quantities
Involve a direction, in addition to numbers and units. (velocity and displacement)
Can be represented graphically with arrows.
The longer the arrow, the greater the magnitude it represents.
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44. Vector Operations Drawing Vectors
In order to draw vectors that indicate direction, you need to work within a coordinate system.
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45. Vector Operations
Coordinate Systems
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46. Vector Operations Coordinate System
When working with the Cartesian coordinates, adding vectors can be accomplished with the “tip-to-tail” method.
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47. Example: A child walks 2. 0 m east, pauses, and then continues 3
Example: A child walks 2.0 m east, pauses, and then continues 3.0 m east.
The resultant (R) = 5.0 m east.

48. If the two vectors have different directions, they are still added tip to tail.
Example: A child walks 2.0 m east, then turns around and walks 4.0 m west.
The resultant is 2.0 m west.

49. Component vectors are added “tip-to-tail.”
The resultant vector is drawn “tail-to-tip.”
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50. Adding vectors graphically, using the “tip-to-tail” method.
4 m east
3 m north
Adding vectors graphically, using the “tip-to-tail” method.
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51. A man walks 3 m north, and then 4 m east. Find his displacement.
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52. 4 m east
3 m north
You are allowed to move the vectors, but don’t change the direction or length.
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53. Line up the tip of one vector with the tail of the other.
4 m east
3 m north
Line up the tip of one vector with the tail of the other.
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54. Line up the tip of one vector with the tail of the other.
4 m east
3 m north
Line up the tip of one vector with the tail of the other.
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55. Now, draw the resultant vector from “tail-to-tip” as shown above.
4 m east
3 m north
Resultant vector
Now, draw the resultant vector from “tail-to-tip” as shown above.
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56. Vector Operations
When we want to add two component vectors that are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant vector, and the tangent function to find the direction of the resultant vector.
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57. Vector Operations 3.02+4.02 = R2 R =  25m2 = 5.0m Resultant vector=R
3.0 m north
4.0 m east
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58. tan = opp/adj tan-1(opp/adj) =  tan-1(3.0m/4.0m) = 37
Vector Operations
tan = opp/adj tan-1(opp/adj) = 
tan-1(3.0m/4.0m) = 37
5.0 m NE
3.0 m north
Full Answer:
5.0m at 37°N of E 
4.0 m east
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59. tan = opp/adj tan-1(opp/adj) = 
What if you know the resultant vector angle and need to find a component vector?
tan = opp/adj tan-1(opp/adj) = 
tan(adj) = opp
Tan37(4.0m)= ? ?
=37
3.01m
4.0 m east
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60. Then use the Pythagorean theorem to find hypotenuse…
3.01m2+4.0m2 = R2 ?
3.01m
R= 25m2=5.0m
=37
4.0 m east
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61. Problems 3A on page 91
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62. Vector Operations
Resolution of Vectors – The process of breaking up a vector into two or more components.
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63. The ball moves forward as it moves upwards and as it falls down.
Vector Operations
Resolution of Vectors
When a golf ball is hit upwards at an angle, it will move in two dimensions.
The ball moves forward as it moves upwards and as it falls down.
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64. Vector Operations Resolution of Vectors
In order to figure out how far the ball will travel, you will be required to resolve the initial velocity vector into horizontal velocity (x) and vertical velocity (y) components.
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65. Vector Operations
Resolution of Vectors
Resolving a vector into its components is the opposite of combining component vectors into a resultant vector.
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66. If you were given component vectors A and B, you could find C.
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67. If you are given C, you need to be able to find Cy and Cx.
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68. Vector Operations
A cannonball is fired with an initial velocity of 35 m/s at an angle of 55o above the horizontal. Vi
55o
55o Vi
V = 35 m/s
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69. Vector Operations
Draw the x and y components of the velocity vector to form a right triangle.
Use Trigonometry functions to find the magnitude of each component. Vi V Vi Vy
55o
55o Vx Vi
V = 35 m/s
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70. The original velocity vector is the hypotenuse of the triangle.
Vector Operations
The original velocity vector is the hypotenuse of the triangle.
IT IS NOT THE PATH OF THE PROJECTILE!!!
hyp Vi Vi V Vy
55o
55o Vx Vi
V = 35 m/s
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71. The vertical component (Vy) represents the opposite side.
Vector Operations
The vertical component (Vy) represents the opposite side.
hyp
opp Vi V Vi Vy
55o
55o Vx Vi
V = 35 m/s
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72. The horizontal component (Vx) represents the adjacent side.
Vector Operations
The horizontal component (Vx) represents the adjacent side.
hyp
opp Vi Vi V Vy
55o
55o Vx
adj Vi
V = 35 m/s
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73. Vector Operations hyp opp Vi From Geometry, we recall opp 55oVy
From Geometry, we recall
opp
Sin  = -----
hyp
55o
55o Vx
adj Vi
V = 35 m/s
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74. Vector Operations We multiply both sides of the equation by hyp hyp
opp Vi Vi V Vy
opp
hyp x Sin  = x hyp
hyp
55o
55o Vx
adj Vi
V = 35 m/s
hyp x sin  = opp
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75. Vector Operations Vy = Visin 
hyp x sin  = opp or
opp = hyp x sin 
hyp
opp Vi Vi V Vy
Remember, opp is really Vy and hyp is Vi, so,
Vy = Visin 
55o Vx
adj
V = 35 m/s Vi
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76. Vector Operations Vy = Visin  hyp Vy = (35 m/s)(sin 55o) opp Vi V VyVx
adj
Vy = 29 m/s Vi
V = 35 m/s
Vy = 29 m/s
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77. Now, for the horizontal (X) component of the initial velocity.
Resolution of Vectors
Now, for the horizontal (X) component of the initial velocity.
From our diagram, we can see that Vx represents the adjacent side of the triangle.
hyp
opp Vi Vi V Vy
55o Vx
adj
V = 35 m/s Vi
Vy = 29 m/s
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78. Resolution of Vectors
We could use either cosine, tangent or the Pythagorean theorem to find the value of Vx.
hyp
opp Vi Vi V Vy
55o Vx
adj
V = 35 m/s Vi
Vy = 29 m/s
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79. Vector Operations Vi 55o adj cos  = ------- hyp hyp
Multiplying both sides by hyp, we get
adj
hyp x cos  = x hyp
hyp
hyp
opp V Vi Vi Vy
55o
55o Vx
adj Vi
V = 35 m/s
Vy = 29 m/s
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80. Vector Operations hyp x cos  = adj, or adj = hyp x cos 
We recall that adj = Vx so,
Vx = Vicos 
Vx = (35 m/s)(cos 55o)
Vx = m/s
Vx = 2.0 X101 m/s
hyp
opp Vi Vi V Vy
55o Vx
adj
V = 35 m/s Vi
Vy = 29 m/s
Vx = 20. m/s
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81. One more thing…
If the angle is below horizontal, or it is a ski slope, etc. draw it like this… Vx  Vy Vi
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82. Problems 3B page 94
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83. Please take out 3B for me to check
Happy Friday!
Please take out 3B for me to check
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84. Adding Vectors Algebraically
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85. Example
A car drives 67 km at an angle of 20 degrees south of east and then drives 78 km at an angle of 67 degrees north of east. What is the car’s resultant displacement ?(Give magnitude and direction – ALWAYS!).
SIX Steps…
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86. 1. Draw a LARGE diagram showing all the vectors(ADDED TIP TO TAIL), all angles, the Resultant Vector and finally all components!
d2=78km R d2
dx1
dy2
20
dy1 d1
67
dx2
d1=67km
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87. 2. Find all the x components and y components of each vector(Remember signs!!).
dy1 = -d1sin 
= -(67km)(sin20)
= -23km
dy2 = d2sin 
(78km)(sin67)
= + 72km
dx1 = d1cos 
= (67km)(cos20)
=+63km
dx2 = d2cos 
=(78km)(cos67)
= + 31km
d2=78km R
dy,2 d2
dx1
20
dy1 d1
67
dx2
d1=67km
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88. 3. Find the total x and total y (Remember signs!!)
dx.Total = 63km+31km=
+94km
dy.total =
-23km+72km=
+49km R d2
dx1
dy2
23
dy1 d1
67
dx2
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89. 4. Draw a NEW triangle showing the total x and y, then draw in the resultant(this should look similar to the 1st resultant you drew if your angles were estimated correctly.) R
yT=49km
xT=94km
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90. 5. Use the Pythagorean Theorem to calculate R…
xT2 + yT2 = R2
94km2+49km2=R2
R2 = 11237km2
R = 106km =1.1X102km
1.1X102km R
49km
94km
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91. You are not finished!!
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92. Full Answer is 1.1X102km at 28 North of East
6. Use the tangent function to determine 
Tan-1(opp/adj) = 
Tan-1(49/94)
= 28  North of East
Full Answer is 1.1X102km at 28 North of East
1.1X102km R
49km 
94km
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93. You should be able to do this with any number of vectors…just a pain…
32
25
53 R
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94. Also, if a direction is already straight up or down or left or right, you do not have to resolve it into components, it is just one component already!!
Ex: If a cat climbs up a tree, it is only a Y component and there is no X component – so call X zero for that vector.
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95. HOMEWORK Problems 3C, page 97
-page 95 has a good example also, if you need it – basically the same problem as my example with different numbers.
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96. Horizontal Projectile Motion
The motion in the x dimension has no effect on the motion in the y dimension, when air resistance is ignored, so “forward” velocity won’t keep objects in the air longer.
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97. The path of a projectile is a curve called a parabola.
Air resistance affects the path (page 99)
Projectile motion is freefall, with an initial horizontal velocity.
See figure 3-19
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98. Horizontal motion is considered constant. Vx,i = Vx,f = constant
Recall Vertical Freefall Formulas from Rest (now the negative is in the formula… don’t ask me why …so put a positive 9.81 m/s2 in these formulas).
Vy,f = -gt
Vy,f2 = -2gy
y = -1/2g(t)2
Horizontal motion is considered constant.
Vx,i = Vx,f = constant
so you can use x = vxt for ANY horizontal distance calculation
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99. Example 1
A softball is thrown horizontally off the top of a 890.0m building. It hits the ground 90.0 m from the base of the building. How long was the ball in the air?
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100. 1. Draw a picture…
Vx (unknown)
890.0 m
90.0 m
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101. 2. Look at possible formulas to find time…
Vx (unknown)
x = vxt
y = -1/2g(t)2
890.0 m
90.0 m
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102. 3. Solve for time with the second formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
890.0 m
t = -2 y/g
90.0 m
t = -(2 ·-890.0)/9.81=13.5 s
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103. Example 2
Michael Franklin is thrown horizontally off the top of a 25m building. He hits the ground* 27.0 m from the base of the building. How fast was Michael thrown?
*of course there was a safety net… 
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104. 12/31/2018 7:09 PM

105. 1. Draw a picture…
Vx (unknown)
25 m
27.0 m
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106. 2. Look at possible formulas to find time…because we can’t get velocity without time and distance!
Vx (unknown)
x = vxt
y = -1/2g(t)2
25 m
27.0 m
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107. 3. Solve for time with the second formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
25 m
t = -2 y/g
27.0 m
t = -(2 ·-25m)/9.81m/s=2.3s
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108. 4. Then solve for vx with the first formula
Vx (unknown)
x = vxt
y = -1/2g(t)2
t = 2.3s
25 m
27.0 m
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109. Vx (unknown) x = vxt y = -1/2g(t)2 25 m t = 2.3s x/ t = vx
Vx=27.0m/ 2.3s = 12m/s
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110. If you know horizontal velocity and distance in the x direction, you could solve for vertical y distance too…
Find time…using
Then find y using
x = vxt
y = -1/2g(t)2
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111. Homework
Problems 3D page 102
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112. Marshmallow Shooter Lab
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113. Projectiles Launched at an Angle
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114. The cart below moves to the right with a uniform velocity of 3. 0 m/s
The cart below moves to the right with a uniform velocity of 3.0 m/s. As it moves, it launches a ball straight up with a velocity of 2.0 m/s. Will the ball land in front of the cart, behind the cart or on top of the cart? (ignore air resistance.)
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115. Answer. In addition to the vertical velocity of 2
Answer. In addition to the vertical velocity of 2.0 m/s, the ball will have the same velocity in the horizontal as the cart. If there is no air resistance, the ball will continue moving forward at 3.0 m/s, until it lands back in the launch tube on the cart.
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116. Components to analyze objects launched at an angle
If the initial velocity vector makes an angle with the horizontal, the motion must be resolved into its components.
Use sine and cosine functions
Vx,i = Vi(cos) and Vy,i = Vi(sin)
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117. Formulas for Projectiles Launched at an angle
Vx= Vi(cos) = constant
x = Vi(cos) t
Vy,f= Vi(sin) - g t
Vy,f2= Vi2(sin)2 – 2g y
y = Vi(sin) t – 1/2g (t)2
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118. Example one Let’s try a complete projectile motion problem.
A golfer hits the golf ball, giving it an initial velocity of 28 m/s, at an angle of 52o above the horizontal. If we ignore air resistance, can you find
How long the ball will be in the air? (t)
How high will it go?(ymax)
How far will it go? (x)
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119. How long the ball will be in the air? (T)
We know that ½ way through it’s parabolic path, the ball will reach a velocity of 0 m/s in the vertical (y) dimension.
In other words, we say at max Y, the Vy,f = 0 m/s.
Vi = 28m/s
Viy
52o
Vix
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
At max Y, the Vy,f = 0 m/s
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120. How long the ball will be in the air? (T)
Vy,f = Vi sin - gt
Because Vy,f = 0 m/s, we can cross it out.
Isolating t we get
t = Vi sin  = 28 m/s sin 52 =
g m/s2
Of course, this really represents 1/2 T, so the ball is in the air 4.4 s.
Vi = 28m/s
Viy
52o
2.2s
Vix
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
At max Y, the Vy,f = 0 m/s
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121. How high will it go?(Y max)
Given
Vi = 28m/s
 = 52
g = 9.81 m/s2
Formula: Vy,f2 = Vi2 (sin )2 - 2g Y
Again, at the max Y , Vy,f = 0 m/s
Isolating Ymax, we get
Vi2 (sin ) m2/s2
Ymax = =
2g (9.81 m/s2)
T = 4.4 s
At max Y, the Vfy = 0 m/s
Find Ymax
=25m
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122. How far will it go? ( X) Given Vi = 28m/s Formula: X = VixT  = 52
g = 9.81 m/s2
Formula: X = VixT
X = VicosT
= (17.24 m/s)(4.4 s) =76 m
T = 4.4 s ymax = 25 m
At max y, the Vy,f = 0 m/s
Find X
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123. A rock is thrown off a cliff with a velocity of 12 m/s at an angle of 34 degrees above horizontal. It lands on the ground after 6.7 seconds. How high does it go? How high is the cliff? and how far does it go?
t=6.7 seconds
 =34º
ymax=?
ycliff = ?
x=?
Example 2
Vx= Vi(cos) = constant
x = Vi(cos) t
Vy,f= Vi(sin) - g t
Vy,f2= Vi2(sin)2 – 2g y
y = Vi(sin) t – 1/2g (t)2
Vyf=0
Vi=12m/s
34º
y=?
x=?
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124. y max Vy,f2= Vi2(sin)2 – 2g y, Vyf=0 ymax= (Vi2(sin)2)/2g
ymax= (Vi2(sin)2)/2g=((12m/s)2sin(34)2)/2(9.81m/s2)=
=2.3m
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125. y cliff y = Vi(sin) t – 1/2g (t)2
y = 12m/s(sin34)6.7s–1/2(9.81m/s2)(6.7s)2
y = -175m, so the cliff is 175m high
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126. x x = Vi(cos) t x = 12m/s(cos34) 6.7s= x = 66.7m
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127. Example 3E on page 103-4 is good too!!
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128. Practice 3E on page 104
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129. Skip 3F…
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130. Relative Velocity
Observers using different frames of reference may measure different displacements or velocities for an object in motion.
Figure 3-23, page 106
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131. Example
A boat heading north crosses a wide river with a velocity of km/h relative to the water. The river has a uniform velocity of 5.00 km/h due east. Determine the boat’s velocity with respect to an observer on shore.
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132. List givens and draw a picture
V (boat/river) = km/h
V (river/shore) = 5.00 km/h
V (boat/shore) = ?
Vr/s
Vb/r
Vb/s 
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133. List givens and draw a picture
V (boat/river) = km/h
V (river/shore) = 5.00 km/h
V (boat/shore) = ?
5.00km/h
10.00km/h
Vb/s 
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134. List givens and draw a picture
Find Vb/s
Pythagorean Theorem
= Vb/s2
Vb/s = 11.18km/h
Find 
Tan-1 (5.00/10.00)
=  = 26.6 East of North
5.00km/h
10.00km/h 
Vb/s
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135. Problems 3F on page 109
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