2. Vectors & Scalars Vectors are measurements which have both magnitude (size) and a directional component. Scalars are measurements which have only magnitude (size) and no directional component EXAMPLES OF VECTOR VALUES: Displacement Velocity Acceleration Force Direction counts in all of these measurements. EXAMPLES OF SCALAR VALUES: Distance Speed Temperature

3. Comparing Vector & Scalar Values Displacement (a vector) versus distance (a scalar) LAKE TRANQUILITY A B We want to get from point A to point B. If we follow the road around the lake our direction is always changing. There is no specific direction. The distance traveled on the road is a scalar quantity. A straight line between A and B is the displacement. It has a specific direction and is therefore a vector.

4. Speed & Velocity Speed and velocity are not the same. Velocity requires a directional component and is therefore a vector quantity. Speed tells us how fast we are going but not which way. Speed is a scalar (direction doesn’t count!) 90 80 70 6050 40 30 20 10 SPEEDOMETER N S EW COMPASS

5. Vectors are typically illustrated by drawing an ARROW  The direction of the physical quantity is given by the direction of the arrow.  The magnitude of the quantity is given by the length of the arrow and it is denoted by |F|.  It is a good habit to label your vectors in a diagram  The direction of the physical quantity is given by the direction of the arrow.  The magnitude of the quantity is given by the length of the arrow and it is denoted by |F|.  It is a good habit to label your vectors in a diagram F=10N VECTOR

6. MAGNITUDE & DIRECTION  MAGNITUDE DISPLACEMENT - METERS, FEET, MILES ETC. FOR VELOCITY - METERS PER SECOND, FEET PER MINUTE, etc FORCE - NEWTONS, DYNES OR POUNDS.  DIRECTION - DEGREES - RADIANS - GEOGRAPHIC INDICATORS SUCH AS NORTH, EAST, SOUTH, NORTHEAST, ETC. - KNOWLEDGE OF COORDINATE GEOMETRY (X-Y PLANE)

7. Up = +Down = -Right = + Left = - y x + + - - Quadrant IQuadrant II Quadrant IIIQuadrant IV 0 o East 90 o North West 180 o 270 o South 360 o Rectangular Coordinates

8. 0 O East 90 O North West 180 O 270 O South 360 O +x +y - x - y 120 O -240 O 30 O West of North 30 O Left of +y 60 O North of West 60 O Above - x MEASURING THE SAME DIRECTION IN DIFFERENT WAYS

9. VECTOR NOTATIONS VECTOR NOTATION MAY TAKE SEVERAL DIFFERENT FORMS: POLAR /CIRCULAR FORM INDICATES A MAGNITUDE VALUE AND A DIRECTIONAL VALUE. THE DIRECTION VALUE MAY BE IN DEGREES, RADIANS OR GEOGRAPHIC TERMS. RECTANGULAR FORM IDENTIFIES THE X-Y COORDINATES OF THE VECTOR. THE VECTOR ITSELF EXTENDS FROM ORIGIN TO THE X- Y POINT. –EXAMPLES: (X = +10, Y = -10) or (10, -10) –THE MAGNITUDE OF THE VECTOR CAN BE FOUND USING THE PYTHAGOREAN THEOREM (10 2 + (-10 2 )) 1/2 = 14.1 - THE DIRECTION CAN BE FOUND USING AN INVERSE TANGENT FUNCTION TAN -1 (10/10) = TAN -1 (1.0) = 45 O SINCE X IS POSITIVE AND Y IS NEGATIVE THE ANGLE IS -45 O AND IS IN QUADRANT IV OR 315 O

10. Circular Notation Circular notation defines a vector by designating the vector’s magnitude |A| and angle θ relative to the +x axis. Using that notation the vector is written:

11. 315 O 0 O East 90 O North West 180 O 270 O South 360 O +x +y - x - y -45 O or 45 O SOUTH OF EAST POLAR COORDINATES 14 METERS, θ=315 O 14 METERS, θ= -45 O 14 METERS, 45 O SOUTH OF EAST CIRCULAR NOTATIONS A=14 m

12. Circular Notation Example: Force vector with magnitude 12 Newtons oriented at 210 degrees with the + x-axis. F = 12 N, θ = 210° F = 12 N, θ = -150° F = 12 N, 30° South of West

13. Circular Notation In this picture, we have a force vector of 4 Newtons oriented along the -x axis. F = 4 N, θ = 180°

14. Circular Notation In the picture above we have 2 vectors C and D. How do we characterize the two vectors? C = 2, θ = 30° D = 4, θ = -50° or D = 4, θ = 310°

15. Circular Notation Exercises: Draw the vectors graphically. 1. A= 6, 30° North of West 2. B= 3, θ = 330 ° 3. C= 20, 45° West of South 4. D= 5, θ = -240 ° 5. E= 8, 60° SE

16. ADDITION/ SUBTRACTION OF VECTORS TWO METHODS 1. GRAPHICAL METHOD (DRAWING) 2. COMPONENT METHOD (MATHEMATICAL). GRAPHICAL ADDITION AND SUBTRACTION REQUIRES THAT EACH VECTOR BE REPRESENTED AS AN ARROW WITH A LENGTH PROPORTIONAL TO THE MAGNITUDE VALUE AND POINTED IN THE PROPER DIRECTION ASSIGNED TO THE VECTOR. (Note: We need ruler and protractor)

17. SCALE = 10 METERS 50 METERS,θ= 0 O 30 METERS θ= 90 O 30 METERS,θ= 45 O VECTOR ARROWS MAY BE DRAWN ANYWHERE ON THE PAGE AS LONG AS THE PROPER LENGTH AND DIRECTION ARE MAINTAINED

18. VECTORS ARE ADDED GRAPHICALLY BY DRAWING EACH VECTOR TO SCALE AND ORIENTED IN THE PROPER DIRECTION. THE VECTOR ARROWS ARE PLACED HEAD TO TAIL. THE ORDER OF PLACEMENT DOES NOT AFFECT THE RESULT (A + B = B + A) THE RESULT OF THE VECTOR ADDITION IS CALLED THE RESULTANT VECTOR. IT IS MEASURED FROM THE TAIL OF THE FIRST VECTOR ARROW TO THE HEAD OF THE LAST ADDED VECTOR ARROW. THE LENGTH OF THE RESULTANT VECTOR ARROW CAN THEN BE MEASURED AND USING THE SCALE FACTOR CONVERTED TO THE CORRECT MAGNITUDE VALUE. THE DIRECTIONAL COMPONENT CAN BE MEASURED USING A PROTRACTOR. GRAPHICAL METHOD

19. A B C D A B C D R A + + + = BCDR ALL VECTORS MUST BE DRAWN TO SCALE & POINTED IN THE PROPER DIRECTION

20. SCALE = 10 METERS Vector B 50 METERS @ 0 O Vector C 30 METERS @ 90 O Vector A 30 METERS @ 45 O A B C Resultant = 9 x 10 = 90 meters Angle is measured at 40 o To add the vectors Place them head to tail

21. IN ALGEBRA, A – B = A + (-B) OR IN OTHER WORDS, ADDING A NEGATIVE VALUE IS ACTUALLY SUBTRACTION. THIS IS ALSO TRUE IN VECTOR SUBTRACTION. IF WE ADD A NEGATIVE VECTOR B TO VECTOR A THIS IS REALLY SUBTRACTING VECTOR B FROM VECTOR A. VECTOR VALUES CAN BE MADE NEGATIVE BY REVERSING THE VECTOR’S DIRECTION BY 180 DEGREES. IF VECTOR A IS 30 METERS DIRECTED AT 45 DEGREES (QUADRANT I), NEGATIVE VECTOR A IS 30 METERS AT 225 DEGREES (QUADRANT II). WORKING WITH VECTORS GRAPHIC SUBTRACTION Vector A 30 METERS @ 45 O Vector - A 30 METERS @ 225 O

22. A B C D A + - - = BCDR A + + ( - ) + ( - ) = BC DR -C-C = -D-D = A -D-D R B -C-C

23. Graphical Method - Examples 1. A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started? 54.5 m, E30 m, E + R= 84.5 m, E Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

24. Graphical Method - Examples 2. You walk 50 meters 60 degrees north of west, then another 85 meters in opposite direction. Calculate your displacement relative to where you started? N E W S 60° R=35, θ=300°

25. COMPONENT METHOD PYTHAGOREAN THEOREM TRIGONOMETRIC FUNCTIONS HORIZONTAL AND VERTICAL COMPONENTS OF THE VECTOR

26.  A B C Sin = A / C  Cos = B / C  Tan  = A / B A C B A B A RIGHT TRIANGLE C  = Arctan (A / B)

27. y x + + - - 0 radians  radians 3/2  radians 2  radians  Quadrant III Quadrant IV Quadrant I Quadrant II Sin Cos Tan  + + +  + - -  - - +  - + -  /2 radians 90 o 0 o 180 o 270 o 360 o

28. A X COMPONENT Y COMPONENT X COMPONENT Y COMPONENT B X COMPONENT C

29. X Y A AxAx Ay  AxAx A = COS  A = SIN Ay   is the angle between the +x-axis and the vector.

30. THE SIGNS OF THE X AND Y COMPONENTS DEPEND ON WHICH QUADRANT THE VECTOR LIES. VECTORS IN QUADRANT I (0 TO 90 DEGREES) HAVE POSITIVE X AND POSITIVE Y VALUES VECTORS IN QUADRANT II (90 TO 180 DEGREES) HAVE NEGATIVE X VALUES AND POSITIVE Y VALUES. VECTORS IN QUADRANT III (180 TO 270 DEGREES) HAVE NEGATIVE X VALUES AND NEGATIVE Y VALUES. VECTORS IN QUADRANT IV (270 TO 360 DEGREES) HAVE POSITIVE X VALUES AND NEGATIVE Y VALUES. VECTOR COMPONENTS

31. WHAT ARE THE X AND Y COMPONENTS OF THE FOLLOWING VECTORS? 1.A =40 m,  = 60 O ? 2. V= 60 m/s,  = 245 0 ? VECTOR COMPONENTS AxAx A = COS  A = SIN Ay 

32. VECTOR COMPONENTS Exercises: Find the x and y components of the following vectors. 1. A= 100 N, 30° North of East 2. B= 30 km, θ = 330 ° 3. C= 20 m, 45° West of South 4. D= 50 m/s, θ = -240 ° 5. E= 80 N, 60° Southeast

33. Analytical Method of Vector Addition 1. Find the x- and y-components of each vector. A x = A cos  =A y = A sin  = B x = B cos  = B y = B sin  = C x = C cos  =C y = C sin  = 2. Sum the x-components. This is the x-component of the resultant. Rx =Rx = 3. Sum the y-components. This is the y-component of the resultant. R y = 4. Use the Pythagorean Theorem to find the magnitude of the resultant vector. R x 2 + R y 2 = R 2

34. ADDING & SUBTRACTING VECTORS USING COMPONENTS Vector A 30 METERS @ 45 O Vector B 50 METERS @ 0 O Vector C 30 METERS @ 90 O ADD THE FOLLOWING THREE VECTORS USING COMPONENTS (1)RESOLVE EACH INTO X AND Y COMPONENTS V = SIN Vy  VxVx V = COS 

35. COMPONENT METHOD C X = 30 COS 90 0 = 0 C Y = 30 SIN 90 0 = 30 A X = 30 COS 45 0 = 21.2 A Y = 30 SIN 45 0 = 21.2 B X = 50 COS 0 0 = 50 B Y = 50 SIN 0 0 = 0 (1)SOLVE EACH INTO X AND Y COMPONENTS

36. (2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR (2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR R x = 21.2 + 50 + 0 = +71.2 R y = 21.2 + 0 + 30 = +51.2 (3) CONSTUCT A NEW RIGHT TRIANGLE USING THE R x AS THE BASE AND R y AS THE OPPOSITE SIDE (3) CONSTUCT A NEW RIGHT TRIANGLE USING THE R x AS THE BASE AND R y AS THE OPPOSITE SIDE R x = +71.2 R y = +51.2 THE HYPOTENUSE IS THE RESULTANT VECTOR R

37. (4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR (4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR R x = +71.2 R y = +51.2 (+71.2) 2 + (+51.2) 2 = 87.7 (5) FIND THE ANGLE (DIRECTION) USING INVERSE TANGENT OF THE OPPOSITE SIDE OVER THE ADJACENT SIDE (5) FIND THE ANGLE (DIRECTION) USING INVERSE TANGENT OF THE OPPOSITE SIDE OVER THE ADJACENT SIDE θ= TAN -1 (51.2/71.2) ANGLE = 35.7 O QUADRANT I R = 87.7 m, θ = 35.7 O

38. SUBTRACTING VECTORS USING COMPONENTS Vector A 30 m @ 45 O Vector C 30 m @ 90 O Vector B 50 m @ 0 O A - + = BCRA + (- ) + = BCR Vector A 30 METERS @ 45 O - B= 50 METERS @ 180 O Vector C 30 METERS @ 90 O

39. 1. SOLVE THE X AND Y COMPONENTS A X = 30COS 45 0 = 21.2 METERS A Y = 30 SIN 45 0 = 21.2 METERS B X = 50 COS 180 0 = - 50 METERS B Y = 50 SIN 180 0 = 0 METERS C X = 30 COS 90 0 = 0 METERS C Y = 30 SIN 90 0 = 30 METERS

40. (2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR R x = SUM OF THE Xs = 21.2 + (-50) + 0 = -28.8 R Y =SUM OF THE Ys = 21.2 + 0 + 30 = +51.2 (3) CONSTUCT A NEW RIGHT TRIANGLE USING THE R x AS THE BASE AND R Y AS THE OPPOSITE SIDE R x = -28.8 R Y = +51.2 THE HYPOTENUSE IS THE RESULTANT VECTOR R

41. (4) USE THE PYTHAGOREAN THEOREM TO THE FIND THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR R x = -28.8 R y = +51.2 ANGLE TAN -1 (51.2/-28.8) ANGLE = -60.6 0 (180 0 –60.6 0 ) = 119.4 0 QUADRANT II (-28.8) 2 + (+51.2) 2 = 58.7 (5) FIND THE ANGLE (DIRECTION) USING INVERSE TANGENT OF THE OPPOSITE SIDE OVER THE ADJACENT SIDE RESULTANT = 58.7 METERS @ 119.4 O R =

42. Example A bear, searching for food wanders 35 meters east then 20 meters north. Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement. 35 m, E 20 m, N 12 m, W 6 m, S - = 23 m, E -= 14 m, N 23 m, E 14 m, N The Final Answer: R 

43. Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. 15 m/s, N 8.0 m/s, W RvRv  The Final Answer :

44. 1.Ehra begins a three-day hiking trip by first walking 25 km, 30° south of east from her base camp. On the second day, she walks 40 km in a direction 60° north of east. On the third day, she walks 35 km, 50° south of west. What is her resultant displacement from base camp? 2.ICADEMY commuter jet departs NAIA and flies 175 km 30° East of North, turns and flies 150 km, 20° North of West, and finally flies 190 km west to its destination. What is its resultant displacement? 3. Find the resultant velocity vector for the following: V1 = 89 km/h, 55° East of South V2 = 46 km/h, 28° South of West V3 = 37 km/h, South