6/26/ :19 AM 1 Introduction to Vectors Scalars and Vectors In Physics, quantities are described as either scalar quantities or vector quantities.

6/26/ :19 AM 2 Introduction to Vectors Scalar Quantities Involve only a magnitude, which includes numbers and units. Examples include distance and speed.

6/26/ :19 AM 3 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.

6/26/ :19 AM 4 Drawing Vectors In order to draw vectors that indicate direction, you need to work within a coordinate system. Vector Operations

6/26/ :19 AM 5 Coordinate Systems Vector Operations

6/26/ :19 AM 6 Coordinate System When working with the Cartesian coordinates, adding vectors can be accomplished with the “tip-to- tail” method. Vector Operations

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

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. 0

6/26/ :19 AM 9 Component vectors are added “tip-to-tail.” The resultant vector is drawn “tail-to-tip.”

6/26/ :19 AM 10 4 m east 3 m north Adding vectors graphically, using the “tip-to- tail” method.

6/26/ :19 AM 11 A man walks 3 m north, and then 4 m east. Find his displacement. 4 m east 3 m north

6/26/ :19 AM 12 You are allowed to move the vectors, but don’t change the direction or length. 4 m east 3 m north

6/26/ :19 AM 13 Line up the tip of one vector with the tail of the other. 4 m east 3 m north tailtip

6/26/ :19 AM 14 Line up the tip of one vector with the tail of the other. 4 m east 3 m north

6/26/ :19 AM 15 Now, draw the resultant vector from “tail- to-tip” as shown above. 4 m east 3 m north Resultant vector

6/26/ :19 AM 16 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. Vector Operations

6/26/ :19 AM = R 2 Vector Operations 4.0 m east 3.0 m north Resultant vector=R R =  25m 2 = 5.0m

6/26/ :19 AM 18 Vector Operations 4.0 m east 3.0 m north 5.0 m NE tan -1 (3.0m/4.0m) = 37  tan  = opp/adj tan -1 (opp/adj) =   Full Answer: 5.0m at 37°N of E

6/26/ :19 AM m east ? tan  = opp/adj tan -1 (opp/adj) =   =37 What if you know the resultant vector angle and need to find a component vector? ? tan  (adj) = opp Tan37(4.0m)= 3.01m

6/26/ :19 AM m east 3.01m  =37 Then use the Pythagorean theorem to find hypotenuse… ? 3.01m m 2 = R 2 R=  25m 2 =5.0m

6/26/ :19 AM 21 Problems 3A on page 91

6/26/ :19 AM 22 Vector Operations Resolution of Vectors – The process of breaking up a vector into two or more components.

6/26/ :19 AM 23 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.

6/26/ :19 AM 24 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.

6/26/ :19 AM 25 Vector Operations Resolving a vector into its components is the opposite of combining component vectors into a resultant vector. Resolution of Vectors

6/26/ :19 AM 26 If you were given component vectors A and B, you could find C.

6/26/ :19 AM 27 If you are given C, you need to be able to find C y and C x.

6/26/ :19 AM 28 Vector Operations A cannonball is fired with an initial velocity of 35 m/s at an angle of 55 o above the horizontal. ViVi 55 o V = 35 m/s ViVi 55 o

6/26/ :19 AM 29 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. V VyVy 55 o VxVx V = 35 m/s ViVi ViVi 55 o ViVi

6/26/ :19 AM 30 Vector Operations The original velocity vector is the hypotenuse of the triangle. IT IS NOT THE PATH OF THE PROJECTILE!!! V VyVy 55 o VxVx V = 35 m/s hyp ViVi ViVi 55 o ViVi

6/26/ :19 AM 31 Vector Operations The vertical component (V y ) represents the opposite side. V VyVy 55 o VxVx V = 35 m/s hyp opp ViVi ViVi 55 o ViVi

6/26/ :19 AM 32 Vector Operations The horizontal component (V x ) represents the adjacent side. V VyVy 55 o VxVx V = 35 m/s hyp opp adj ViVi ViVi 55 o ViVi

6/26/ :19 AM 33 Vector Operations V VyVy 55 o VxVx V = 35 m/s hyp opp adj From Geometry, we recall opp Sin  = hyp ViVi ViVi 55 o ViVi

6/26/ :19 AM 34 Vector Operations We multiply both sides of the equation by hyp V VyVy 55 o VxVx V = 35 m/s hyp opp adj opp hyp x Sin  = x hyp hyp hyp x sin  = opp ViVi ViVi 55 o ViVi

6/26/ :19 AM 35 Vector Operations V VyVy 55 o VxVx V = 35 m/s hyp opp adj hyp x sin  = opp or opp = hyp x sin  Remember, opp is really V y and hyp is V i, so, V y = V i sin  ViVi ViVi ViVi

6/26/ :19 AM 36 Vector Operations V VyVy 55 o VxVx V = 35 m/s hyp opp adj V y = V i sin  V y = (35 m/s)(sin 55 o) V y = m/s V y = 29 m/s ViVi ViVi ViVi

6/26/ :19 AM 37 Resolution of Vectors Now, for the horizontal (X) component of the initial velocity. From our diagram, we can see that V x represents the adjacent side of the triangle. V VyVy 55 o VxVx V = 35 m/s V y = 29 m/s adj hyp opp ViVi ViVi ViVi

6/26/ :19 AM 38 Resolution of Vectors We could use either cosine, tangent or the Pythagorean theorem to find the value of V x. V VyVy 55 o VxVx V = 35 m/s V y = 29 m/s adj hyp opp ViVi ViVi ViVi

6/26/ :19 AM 39 Vector Operations adj cos  = hyp Multiplying both sides by hyp, we get adj hyp x cos  = x hyp hyp V VyVy 55 o VxVx V = 35 m/s V y = 29 m/s adj hyp opp ViVi ViVi 55 o ViVi

6/26/ :19 AM 40 Vector Operations hyp x cos  = adj, or adj = hyp x cos  We recall that adj = V x so, V x = V i cos  V x = (35 m/s)(cos 55 o) V x = m/s V x = 2.0 X10 1 m/s V VyVy 55 o VxVx V = 35 m/s V y = 29 m/s adj hyp opp V x = 20. m/s ViVi ViVi ViVi

6/26/ :19 AM 41 One more thing… If the angle is below horizontal, or it is a ski slope, etc. draw it like this…  Vx Vy Vi

6/26/ :19 AM 42 Problems 3B page 94

6/26/ :19 AM 43 Adding Vectors Algebraically

6/26/ :19 AM 44 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…

6/26/ :19 AM  67  d1d1 d2d2 d x2 d x1 d y2 d y1 R 1. Draw a LARGE diagram showing all the vectors(ADDED TIP TO TAIL), all angles, the Resultant Vector and finally all components! d 2 =78km d 1 =67km

6/26/ :19 AM  67  d y1 = -d 1 sin  = -(67km)(sin20  ) = -23km d y2 = d 2 sin  (78km)(sin67  ) = + 72km d x1 = d 1 cos  = (67km)(cos20  ) =+63km d x2 = d 2 cos  =(78km)(cos67  ) = + 31km d1d1 d2d2 d x2 d x1 d y,2 d y1 R 2. Find all the x components and y components of each vector(Remember signs!!). d 1 =67km d 2 =78km

6/26/ :19 AM  67  d x.Total = 63km+31km= +94km d y.total = -23km+72km= +49km d1d1 d2d2 d x2 d x1 d y2 d y1 R 3. Find the total x and total y (Remember signs!!)

6/26/ :19 AM 48 x T =94km y T =49km R 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.)

6/26/ :19 AM 49 R 5. Use the Pythagorean Theorem to calculate R… x T 2 + y T 2 = R 2 94km 2 +49km 2 =R 2 R 2 = 11237km 2 R = 106km =1.1X10 2 km 49km 94km 1.1X102km

6/26/ :19 AM 50 You are not finished!!

6/26/ :19 AM 51 R 6. Use the tangent function to determine  Tan-1(opp/adj) =  Tan-1(49/94) = 28  North of East Full Answer is 1.1X10 2 km at 28  North of East 49km 94km  1.1X10 2 km

6/26/ :19 AM  25  32  You should be able to do this with any number of vectors…just a pain… R

6/26/ :19 AM 53 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.

6/26/ :19 AM 54 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.

6/26/ :19 AM 55 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.

6/26/ :19 AM 56 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

6/26/ :19 AM 57 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/s 2 in these formulas). V y,f = -g  t V y,f 2 = -2g  y  y = -1/2g(  t) 2 Horizontal motion is considered constant. V x,i = V x,f = constant so you can use  x = v x  t for ANY horizontal distance calculation

6/26/ :19 AM 58 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?

6/26/ :19 AM Draw a picture… 90.0 m Vx (unknown) m

6/26/ :19 AM Look at possible formulas to find time… 90.0 m Vx (unknown) m  x = v x  t  y = -1/2g(  t)2

6/26/ :19 AM Solve for time with the second formula 90.0 m Vx (unknown) m  x = v x  t  y = -1/2g(  t) 2  t =  -2  y/g  t =  -(2 · )/9.81=13.5 s

6/26/ :19 AM 62 Example 2 Dakota 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 Dakota thrown? *of course there was a safety net

6/26/ :19 AM 63

6/26/ :19 AM Draw a picture… 27.0 m Vx (unknown) 25 m

6/26/ :19 AM m Vx (unknown) 25 m  x = v x  t  y = -1/2g(  t)2 2. Look at possible formulas to find time…because we can’t get velocity without time and distance!

6/26/ :19 AM Solve for time with the second formula 27.0 m Vx (unknown) 25 m  x = v x  t  y = -1/2g(  t) 2  t =  -2  y/g  t =  -(2 · -25m)/9.81m/s=2.3s

6/26/ :19 AM Then solve for v x with the first formula 27.0 m Vx (unknown) 25 m  x = v x  t  y = -1/2g(  t) 2  t = 2.3s

6/26/ :19 AM m V x (unknown) 25 m  x = v x  t  y = -1/2g(  t) 2  t = 2.3s  x/  t = v x V x =27.0m/ 2.3s = 12m/s

6/26/ :19 AM 69 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 = v x  t  y = -1/2g(  t) 2

6/26/ :19 AM 70 Homework Problems 3D page 102

6/26/ :19 AM 71 Projectiles Launched at an Angle

6/26/ :19 AM 72 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.)

6/26/ :19 AM 73 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.

6/26/ :19 AM 74 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 V x,i = V i (cos  ) and V y,i = V i (sin  )

6/26/ :19 AM 75 Formulas for Projectiles Launched at an angle V x = V i (cos  ) = constant  x = V i (cos  )  t V y,f = V i (sin  ) - g  t V y,f 2 = V i 2 (sin  ) 2 – 2g  y  y = V i (sin  )  t – 1/2g (  t) 2

6/26/ :19 AM 76 Example one A golfer hits the golf ball, giving it an initial velocity of 28 m/s, at an angle of 52 o 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?(  y) How far will it go? (  x) Let’s try a complete projectile motion problem.

6/26/ :19 AM 77 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 V y,f = 0 m/s. V i = 28m/s 52 o V iy V ix Given V i = 28m/s  = 52  g = 9.81 m/s 2 At max  Y, the V y,f = 0 m/s

6/26/ :19 AM 78 How long the ball will be in the air? (  T) V y,f = V i sin  - g  t Because V y,f = 0 m/s, we can cross it out. Isolating  t we get  t = V i sin  = 28 m/s sin 52 = g 9.81 m/s 2 Of course, this really represents 1/2  T, so the ball is in the air 4.4 s. V i = 28m/s 52 o V iy V ix At max  Y, the V y,f = 0 m/s 2.2s Given V i = 28m/s  = 52  g = 9.81 m/s 2

6/26/ :19 AM 79 How high will it go?(  Y )  T = 4.4 s At max  Y, the V fy = 0 m/s Find  Y Formula: V y,f 2 = V i 2 (sin  ) 2 - 2g  Y Again, at the max.  Y, V y,f = 0 m/s Isolating  Y, we get Vi 2 (sin  ) m 2 /s 2  Y = = = 25 m 2g 2(9.81 m/s 2 ) Given V i = 28m/s  = 52  g = 9.81 m/s 2

6/26/ :19 AM 80 How far will it go? (  X)  T = 4.4 s  y = 25 m At max  y, the V y,f = 0 m/s Find  X Formula:  X = V ix  T  X = V i cos  T = (17.24 m/s)(4.4 s) =76 m Given V i = 28m/s  = 52  g = 9.81 m/s 2

6/26/ :19 AM 81 Example 3E is good too!!

6/26/ :19 AM 82 Practice 3E on page 104

6/26/ :19 AM 83 Skip 3F…

6/26/ :19 AM 84 Relative Velocity Observers using different frames of reference may measure different displacements or velocities for an object in motion. – Figure 3-23, page 106

6/26/ :19 AM 85 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.

6/26/ :19 AM 86 List givens and draw a picture V (boat/river) = km/h V (river/shore) = 5.00 km/h V (boat/shore) = ? V b/r V r/s V b/s 

6/26/ :19 AM 87 List givens and draw a picture V (boat/river) = km/h V (river/shore) = 5.00 km/h V (boat/shore) = ? 10.00km/h 5.00km/h V b/s 

6/26/ :19 AM 88 List givens and draw a picture Find V b/s Pythagorean Theorem = V b/s 2 V b/s = 11.18km/h Find  Tan-1 (5.00/10.00) =  = 26.6  East of North 10.00km/h 5.00km/h V b/s 

6/26/ :19 AM 89 Problems 3F on page 109