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1 Slide 1 / 246 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website:

2 Slide 2 / 246 Kinematics in Two Dimensions

3 Slide 3 / 246 Table of Contents: Kinematics in 2D Click on the topic to go to that section Kinematics in One Dimension (Review) Adding Vectors in Two Dimensions Basic Vector Operations Vector Components Projectile Motion General Problems

4 Slide 4 / 246 Key Terms and Equations Kinematics Equations: 2-Dimensional Equations v y = v sin(θ) θ = tan -1 (v y / v x ) v = v 0 + a t v x = v cos(θ) v 2 = v a Δx x = x 0 + v 0 t + 1/2 a t 2 v = (v 2 x + v y2 ) Right Triangle Equations: a 2 + b 2 = c 2 SOH CAH TOA sin(θ) = Opposite/Hypotenuse cos(θ) = Adjacent/Hypotenuse tan(θ) = Opposite/Adjacent

5 Slide 5 / 246 Kinematics in One Dimension Return to Table of Contents

6 Slide 6 / 246 Review of 1-D Kinematics Kinematics is the description of how objects move with respect to a defined reference frame. Displacement is the change in position of an object. Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. Instantaneous velocity is the limit as the time becomes infinitesimally short. Average acceleration is the change in velocity divided by the time.

7 Slide 7 / 246 Review of 1-D Kinematics Instantaneous acceleration is the limit as the time interval becomes infinitesimally small. There are four equations of motion for constant acceleration, each requires a different set of quantities. v = v o + at x = x o + v o t + ½at 2 v 2 = v o 2 + 2a(x - x o ) v = v + v o 2

8 Slide 8 / Starting from rest, you accelerate at 4.0 m/s 2 for 6.0s. What is your final velocity? v = v o + at v = 0 + 4(6) v = 24m/s / njc.tl/ 3i

9 Slide 9 / You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s 2 for 9.0s; what is your final velocity? v = v o + at v = (9) v = 19.5m/s / njc.tl/ 3j

10 Slide 10 / How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is -2.0 m/s 2? v = v o + at 0 = t t = 2.5s / njc.tl/ 3k

11 Slide 11 / An object moves at a constant speed of 6 m/s. This means that the object: A B C D E Increases its speed by 6 m/s every second Decreases its speed by 6 m/s every second Doesn t move Has a positive acceleration Moves 6 meters every second / njc.tl/ 3l

12 Slide 12 / A snapshot of three racing cars is shown on the diagram. All three cars start the race at the same time, at the same place and move along a straight track. As they approach the finish line, which car has the lowest average speed? A B C D E Car I Car II Car III All three cars have the same average speed More information is required / njc.tl/ 3m

13 Slide 13 / 246 Motion at Constant Acceleration In physics there is another approach in addition to algebraic which is called graphical analysis. The following formula v = v 0 + at can be interpreted by the graph. We just need to recall our memory from math classes where we already saw a similar formula y = mx + b.

14 Slide 14 / 246 Motion at Constant Acceleration From these two formulas we can find some analogies: Velocity v y (depended variable of x), v 0 b (intersection with vertical axis), t x (independent variable), a m ( slope of the graph- the ratio between rise and run Δy/Δx). y = m x + b v = a t + v 0 (or v = v 0 + a t)

15 Slide 15 / 246 Motion at Constant Acceleration The formula a =Δv/Δt shows that the value of acceleration is the same as the slope on a graph of velocity versus time.

16 Slide 16 / The velocity as a function of time is presented by the graph. What is the acceleration? a = slope = Δv/Δt = (10 m/s -2 m/s)/40 s = 0.2 m/s 2 / njc.tl/ 3n

17 Slide 17 / The velocity as a function of time is presented by the graph. Find the acceleration. a = slope = Δv/Δt = (0 m/s - 25 m/s)/10 s = -2.5 m/s 2 / njc.tl/ 3o

18 Slide 18 / 246 Motion at Constant Acceleration The acceleration graph as a function of time can be used to find the velocity of a moving object. When the acceleration is constant it can be shown on the graph as a straight horizontal line.

19 Slide 19 / 246 Motion at Constant Acceleration In order to find the change in velocity for a certain limit of time we need to calculate the area under the acceleration versus time graph. The change in velocity during first 12 seconds is equivalent to the shadowed area (4χ12 = 48). The change in velocity during first 12 seconds is 48 m/s.

20 Slide 20 / What is the velocity of the object at 5 s? A B C D E 1 m/s 2 m/s 3 m/s 4 m/s 5 m/s / njc.tl/ 3p

21 Slide 21 / Which of the following statements is true? A B C D E The object speeds up The object slows down The object moves with a constant velocity The object stays at rest The object is in free fall / njc.tl/ 3q

22 Slide 22 / The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds? Δv = area = (3χ10) = 30 m/s / njc.tl/ 3r

23 Slide 23 / The graph represents the relationship between velocity and time for an object moving in a straight line. What is the traveled distance of the object at 9 s? A B C D E 10 m 24 m 36 m 48 m 56 m / njc.tl/ 3s

24 Slide 24 / Which of the following is true? A B C D E The object increases its velocity The object decreases its velocity The object s velocity stays unchanged The object stays at rest More information is required / njc.tl/ 3t

25 Slide 25 / What is the initial position of the object? A B C D E 2 m 4 m 6 m 8 m 10 m / njc.tl/ 3u

26 Slide 26 / What is the velocity of the object? A B C D E 2 m/s 4 m/s 6 m/s 8 m/s 10 m/s / njc.tl/ 3v

27 Slide 27 / What is the initial position of the object? A B C D E 2 m 4 m 6 m 8 m 10 m / njc.tl/ 3u

28 Slide 28 / The graph represents the position as a function of time of a moving object. What is the velocity of the object? A B C D E 5 m/s -5 m/s 10 m/s -10 m/s 0 m/s / njc.tl/ 3w

29 Slide 29 / 246 Free Fall All unsupported objects fall towards the earth with the same acceleration. We call this acceleration the "acceleration due to gravity" and it is denoted by g. g = 9.8 m/s 2 Keep in mind, ALL objects accelerate towards the earth at the same rate. g is a constant!

30 Slide 30 / 246 What It stops happens momentarily. at the top? v = 0 g = -9.8 m/s 2 It slows down. What happens when it (negative acceleration) goes up? g = -9.8 m/s 2 It speeds up What happens when it (negative goes down? acceleration) g = -9.8 m/s 2 An object is thrown upward with initial velocity, v o It What returns happens with when its it original lands? velocity.

31 Slide 31 / 246 It stops momentarily. v = 0 g = -9.8 m/s 2 It slows down. (negative acceleration) g = -9.8 m/s 2 It speeds up. (negative acceleration) g = -9.8 m/s 2 An object is thrown upward with initial velocity, v o It returns with its original velocity.

32 Slide 32 / 246 On the way up: v a t = 3 s On the way down: v 0 v 1 a t = 0 s v 2 a t = 2 s a v 2 v 1 a t = 1 s v 1 v 2 t = 1 s a v v 2 t = 2 s v 0 v 1 t = 0 s a v t = 3 s

33 v (m/s) Slide 33 / 246 For any object thrown straight up into the air, what does the velocity vs time graph look like? An object is thrown upward with initial velocity, v o It stops momentarily. v = 0 g = -9.8 m/s 2 t (s) It returns with its original velocity but in the opposite direction.

34 Slide 34 / An object moves with a constant acceleration of 5 m/s 2. Which of the following statements is true? A B C D E The object s velocity stays the same The object moves 5 m each second The object s acceleration increases by 5 m/s2 each second The object s acceleration decreases by 5 m/s2 each second the object s velocity increases by 5 m/s each second / njc.tl/ 3x E

35 Slide 35 / A truck travels east with an increasing velocity. Which of the following is the correct direction of the car s acceleration? A B C D E A

36 Slide 36 / A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the final speed of the car compared to the truck? A B C D E Half as much The same Twice as much Four times as much One quarter as much / njc.tl/ 3y

37 Slide 37 / A car and a delivery truck both start from rest and accelerate at the same rate. However, the car accelerates for twice the amount of time as the truck. What is the traveled distance of the car compared to the truck? A B C D E Half as much The same Twice as much Four times as much One quarter as much

38 Slide 38 / A modern car can develop an acceleration four times greater than an antique car like Lanchester If they accelerate over the same distance, what would be the velocity of the modern car compared to the antique car? A B C D E Half as much The same Twice as much Four times as much One quarter as much / njc.tl/ 40 C

39 Slide 39 / An object is released from rest and falls in the absence of air resistance. Which of the following is true about its motion? A B C D E Its acceleration is zero Its acceleration is constant Its velocity is constant Its acceleration is increasing Its velocity is decreasing B / njc.tl/ 41

40 Slide 40 / A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction of the ball s velocity and acceleration between A and B? A D B E C B / njc.tl/ 42

41 Slide 41 / A ball is thrown straight up from point A it reaches a maximum height at point B and falls back to point C. Which of the following is true about the direction the ball s velocity and acceleration between B and C? A D B E C D / njc.tl/ 43

42 Slide 42 / A football, a hockey puck, and a tennis ball all fall down in the absence of air resistance. Which of the following is true about their acceleration? A B C D E The acceleration of the football is greater than the other two The acceleration of the hockey puck is greater than the other two The acceleration of the tennis ball is greater than the other two They all fall down with the same constant acceleration More information is required / njc.tl/ 44 D

43 Slide 43 / A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is H and in the second trial 4H. Compare the time it takes for the package to reach the surface in the second trial to that in the first trial? A B C D E The time in the second trial is four times greater The time in the second trial is two times greater The time the same in both trials because it doesn t depend on height The time in the second trial is four times less The time in the second trial is two times less / njc.tl/ 45 B

44 Slide 44 / Two baseballs are thrown from the roof of a house with the same initial speed, one is thrown up, and the other is down. Compare the speeds of the baseballs just before they hit the ground. A The one thrown up moves faster because the initial velocity is up B C D E The one thrown down moves faster because the initial velocity is down They both move with the same speed The one thrown up moves faster because it has greater acceleration The one thrown down moves faster because it has greater acceleration / njc.tl/ 46 B

45 Slide 45 / An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is v0 and second time he increases the initial sped to 4v0. How would you compare the maximum height in the first trial to that in the first trial? A Two times greater B Four times greater C Eight times greater D Sixteen times greater E The same / njc.tl/ 47 D

46 Slide 46 / An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground? v = v o + at v=0-(9.8)2.5 v=-24.5 m/s / njc.tl/ 48

47 Slide 47 / An arrow is fired into the air and it reaches its highest point 3 seconds later. What was its velocity when it was fired? v = v o + at 0 = v o (3) v o = 29.4m/s / njc.tl/ 49

48 Slide 48 / You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration? v 2 = v o 2 + 2aΔx 60 2 = a(200) a = 8m/s 2 / njc.tl/ 4a

49 Slide 49 / A dropped ball falls 8.0m; what is its final velocity? v 2 = v o 2 + 2aΔy v 2 = 2aΔy v 2 = 2(9.8)8 v = -12.5m/s / njc.tl/ 4b

50 Slide 50 / A ball with an initial velocity of 25m/s is subject to an acceleration of -9.8 m/s 2 ; how high does it go before coming to a momentary stop? v 2 = v o 2 + 2aΔy 0 = (-9.8)Δy y = 31.9m / njc.tl/ 4c

51 Slide 51 / A water drop falls down from the roof of a house during 3 s, how high is the house? x = 1/2 (9.8) (3) 2\ x = 44 m / njc.tl/ 4d

52 Slide 52 / A ball is thrown vertically down from the edge of a cliff with a speed of 8 m/s, how high is the cliff, if it took 6 s for the ball to reach the ground? x = (8)(60 + 1/2(9.8)(6) 2 x = m / njc.tl/ 4e

53 Slide 53 / What is the landing velocity of an object that is dropped from a height of 49 m? x = 31 m/s / njc.tl/ 4f

54 Slide 54 / An object is thrown vertically up with a velocity of 35 m/s. What was the maximum height it reached? x = 62.5 m / njc.tl/ 4g

55 Slide 55 / A boy throws a ball vertically up and catches it after 3 s. What height did the ball reach? x = 11 m / njc.tl/ 4h

56 Slide 56 / 246 Vectors vs. Scalars We used vectors last year.you'll recall that: Vectors have a magnitude AND a direction Scalars only have a magnitude

57 Slide 57 / Which one of the following is an example of a vector quantity? A B C D distance velocity mass area / njc.tl/ 4i

58 Slide 58 / Which of the following is a vector quantity? A B C D E Speed Time Traveled distance Velocity Area / njc.tl/ 4j

59 Slide 59 / A runner runs halfway around a circular path of radius 10 m. What is the displacement of the jogger? A B C D 0 m 10 m 20 m 31.4 m / njc.tl/ 4k

60 Slide 60 / A runner runs halfway around a circular path of radius 10 m. What is the total traveled distance of the jogger? A B C D 0 m 10 m 20 m 31.4 m / njc.tl/ 4l

61 Slide 61 / 246 Adding Vectors in Two Dimensions Return to Table of Contents

62 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 62 / 246 Adding Vectors

63 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 63 / 246 Adding Vectors 1. Draw the first vector, beginning at the origin, with its tail at the origin.

64 Slide 64 / 246 Adding Vectors Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. 1. Draw the first vector, beginning at the origin, with its tail at the origin. 2. Draw the second vector with its tail at the tip of the first vector.

65 Last year, we learned how to add vectors along a single axis. The example we used was for adding two displacements. Slide 65 / 246 Adding Vectors 1. Draw the first vector, beginning at the origin, with its tail at the origin. 2. Draw the second vector with its tail at the tip of the first vector. 3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

66 The direction of each vector matters. Slide 66 / 246 Adding Vectors In this first case, the vector sum of: 5 units to the East plus 2 units to West is 3 units to the East.

67 The direction of each vector matters. For instance, if the second vector had been 2 units to the EAST (not west), we get a different answer. Slide 67 / 246 Adding Vectors In this second case, the vector sum of: 5 units to the East plus 2 units to EAST is 7 units to the East.

68 But how about if the vectors are along different axes. Slide 68 / 246 Adding Vectors in 2-D For instance, let's add vectors of the same magnitude, but along different axes. What is the vector sum of: 5 units East plus 2 units North

69 1. Draw the first vector, beginning at the origin, with its tail at the origin. Slide 69 / 246 Adding Vectors

70 1. Draw the first vector, beginning at the origin, with its tail at the origin. Slide 70 / 246 Adding Vectors 2. Draw the second vector with its tail at the tip of the first vector.

71 1. Draw the first vector, beginning at the origin, with its tail at the origin. Slide 71 / 246 Adding Vectors 2. Draw the second vector with its tail at the tip of the first vector. 3. Draw the Resultant (the answer) from the tail of the first vector to the tip of the last.

72 Slide 72 / 246 Adding Vectors Drawing the Resultant is the same as we did last year. But calculating its magnitude and direction require the use of right triangle mathematics. c a b We know the length of both SIDES of the triangle (a and b), but we need to know the length of the HYPOTENUSE (c).

73 The magnitude of the resultant is equal to the length of the vector. Slide 73 / 246 Magnitude of A Resultant We get the magnitude of the resultant from the Pythagorean Theorem: c =? b = 2 units c 2 = a 2 + b 2 a = 5 units or in this case: R 2 = R 2 = R 2 = 29 R = (29) = 5.4 units

74 Slide 74 / Starting from the origin, a person walks 6 km east during first day, and 3 km east the next day. What is the net displacement of the person from the initial point in two days? A B C D E 6 km, west 3 km, east 10 km, east 5 km, west 9 km, east / njc.tl/ 4m

75 Slide 75 / Starting from the origin, a car travels 4 km east and then 7 km west. What is the net displacement of the car from the initial point? A B C D E 3 km, west 3 km, east 4 km, east 7 km, west 7 km east / njc.tl/ 4n

76 Slide 76 / Starting from the origin, a person walks 8 km east during first day, and 5 km west the next day. What is the net displacement of the person from the initial point in two days? A B C D E 6 km, east 3 km, east 10 km, west 5 km, west 9 km, east / njc.tl/ 4o

77 Slide 77 / What is the magnitude of the Resultant of two vectors A and B, if A = 8.0 units north and B = 4.5 units east? 9.18 units / njc.tl/ 4p

78 Slide 78 / What is the magnitude of the Resultant of two vectors A and B, if A = 24.0 units east and B = 15.0 units south? 28.3 units / njc.tl/ 4q

79 In physics, we say the direction of a vector is equal to the angle θ between a chosen axis and the resultant. Slide 79 / 246 Adding Vectors In Kinematics, we will primarily use the x-axis to measure θ. θ However, if we were to change the axis we used and apply the proper mathematical techniques, we should get the same result!

80 Slide 80 / 246 Adding Vectors To find the value of the angle θ, we need to use what we already know: the length of the two sides opposite and adjacent to the angle. θ (remember SOH CAH TOA) tan(θ) = opposite adjacent

81 Slide 81 / 246 Adding Vectors tan(θ) = opp / adj tan(θ) = (2 units) / (5 units) tan(θ) = 2/5 tan(θ) = 0.40 θ To find the value of θ, we must take the inverse tangent: θ = tan -1 (0.40) = 22 0

82 Slide 82 / 246 Adding Vectors θ The Resultant is 5.4 units in the direction of 22 0 North of East magnitude direction

83 Slide 83 / What is the direction of the Resultant of the two vectors A and B if: A = 8.0 units north and B = 4.5 units east? θ = 60 o / njc.tl/ 4r

84 Slide 84 / What is the direction of the Resultant of the two vectors A and B if: A = 24.0 units east and B = 15.0 units south? θ = 32 o / njc.tl/ 4s

85 Slide 85 / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Magnitude =? 471 units / njc.tl/ 4t

86 Slide 86 / Find the magnitude and direction of the resultant of two vectors A and B if: A = 400 units north B = 250 units east Direction =? θ = 58 o / njc.tl/ 4t

87 Slide 87 / A student walks a distance of 300 m East, then walks 400 m North. What is the magnitude of the net displacement? A B C D 300 m 400 m 500 m 700 m / njc.tl/ 4u

88 Slide 88 / A student walks a distance of 300 m East, then walks 400 m North. What is the total traveled distance? A B C D 300 m 400 m 500 m 700 m / njc.tl/ 4v

89 Slide 89 / Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum A B C D is 2.0 m could be as small as 2.0 m, or as large as 12 m is 12 m is larger than 12 m / njc.tl/ 4w

90 Slide 90 / The resultant of two vectors is the largest when the angle between them is A 0 B 45 C 90 D / njc.tl/ 4x

91 56 The resultant of two vectors is the smallest when the angle between them is Slide 91 / 246 A 0 B 45 C 90 D / njc.tl/ 4y

92 Slide 92 / 246 Basic Vector Operations Return to Table of Contents

93 Slide 93 / 246 Adding Vectors Adding Vectors in the opposite order gives the same resultant. V 1 + V 2 = V 2 + V 1 V 1 V 2 V 2 V 1

94 Slide 94 / 246 Adding Vectors Even if the vectors are not at right angles, they can be added graphically by using the "tail to tip" method. The resultant is drawn from the tail of the first vector to the tip of the last vector. V V V 2 + = V 1 V 2 V R V 3

95 Slide 95 / 246 Adding Vectors...and the order does not matter. V 3 V 2 V 3 V 1 V 2 + V 1 + = V R Try it. V 3 V 1 V 2 V R

96 Slide 96 / 246 Adding Vectors Vectors can also be added using the parallelogram method. V 1 + = V 2 V 1 V R V 2

97 Slide 97 / 246 Subtracting Vectors In order to subtract a vector, we add the negative of that vector. The negative of a vector is defined as that vector in the opposite direction. V 1 - = V 2 V 1 + -V 2 = V R -V 2 V 1

98 Slide 98 / 246 Multiplication of Vectors by Scalars A vector V can be multiplied by a scalar c. The result is a vector cv which has the same direction as V. However, if c is negative, it changes the direction of the vector. V 2V -½V

99 57 Which of the following operations will not change a vector? Slide 99 / 246 A B C D Translate it parallel to itself Rotate it Multiply it by a constant factor Add a constant vector to it / njc.tl/ 4z

100 Slide 100 / 246 Vector Components Return to Table of Contents

101 Slide 101 / 246 Adding Vectors by Components Any vector can be described as the sum as two other vectors called components. These components are chosen perpendicular to each other and can be found using trigonometric functions. y V y V θ V x x

102 Slide 102 / 246 Adding Vectors by Components In order to remember the right triangle properties and to better identify the functions, it is often convenient to show these components in different arrangements (notice v y below). y V V y θ V x x

103 Slide 103 / Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The y component of A is A B C D 6.9 m -6.9 m 4.0 m -4.0 m / njc.tl/ 50

104 Slide 104 / Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis. The X component of A is A B C D 6.9 m -6.9 m 4.0 m -4.0 m / njc.tl/ 50

105 Slide 105 / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the vertical component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s / njc.tl/ 51

106 Slide 106 / If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the horizontal component of the velocity? A B C D 12 m/s 15 m/s 20 m/s 25 m/s / njc.tl/ 51

107 Slide 107 / If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked? A 19 B 45 C 60 D 71 / njc.tl/ 52

108 Slide 108 / A butterfly moves with a speed of 12.0 m/s. The x component of its velocity is 8.00 m/s. The angle between the direction of its motion and the x axis must be A B C D / njc.tl/ 53

109 Slide 109 / A 400-m tall tower casts a 600-m long shadow over a level ground. At what angle is the Sun elevated above the horizon? A 34 B 42 C 48 D can't be found; not enough information / njc.tl/ 54

110 Slide 110 / 246 Adding Vectors by Components Using the tip-to-tail method, we can sketch the resultant of any two vectors. y V V 2 V 1 But we cannot find the magnitude of 'v', the resultant, since v 1 and v 2 are two-dimensional vectors x

111 Slide 111 / 246 Adding Vectors by Components We now know how to break v 1 and v 2 into components... y V 2 V 2y V 1 V 2x V 1y V 1x x

112 Slide 112 / 246 Adding Vectors by Components And since the x and y components are one dimensional, they can be added as such. y V 2y V y = v 1y + v 2y V 1y V 1x V 2x V x = v 1x + v 2x x

113 Slide 113 / Two vectors A and B have components (0, 1) and (-1, 3), respectively. What are the components of the sum of these two vectors? A (1, 4) B (-1, 4) C (1, 2) D (-1, 2) / njc.tl/ 55

114 Slide 114 / Two vectors A and B have components (0, 1) and (-1, 3), respectively. What is magnitude of the sum of these two vectors? A 2.8 B 3.2 C 3.9 D / njc.tl/ 56

115 Slide 115 / Vector A = (1, 3). Vector B = (3, 0). Vector C = A + B. What is the magnitude of C? A 3 B 4 C 5 D 7 / njc.tl/ 57

116 Slide 116 / 246 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2 V 1

117 y Slide 117 / 246 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2 2. Choose x and y axes. V 1 x

118 y Slide 118 / 246 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2 V 2y 2. Choose x and y axes. V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x

119 y Slide 119 / 246 Adding Vectors by Components 1. Draw a diagram and add the vectors graphically. V 2 V 2y 2. Choose x and y axes. V 1 V 1y V 2x 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. v 1x = v 1 cos(θ 1 ) v 2x = v 2 cos(θ 2 ) v 1y = v 1 sin(θ 1 ) v 2y = v 2 sin(θ 2 )

120 y Slide 120 / 246 Adding Vectors by Components V x 1. Draw a diagram and add the vectors graphically. V y V 1 V 2 V 1y V 2x V 2y 2. Choose x and y axes. 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. 5. Add the components in each direction.

121 y Slide 121 / 246 Adding Vectors by Components V x 1. Draw a diagram and add the vectors graphically. V y V V 1 V 2 V 1y V 2x V 2y 2. Choose x and y axes. 3. Resolve each vector into x and y components. V 1x x 4. Calculate each component. v = (v x 2 + v y2 ) θ = tan -1 ( v y ) v x (length) (direction) 5. Add the components in each direction. 6. Find the length and direction of the resultant vector.

122 Slide 122 / 246 Example: Graphically determine the resultant of the following three vector displacements: 1. 24m, 30º north of east 2. 28m, 37º east of north 3. 20m, 50º west of south

123 Slide 123 / m, 30º north of east Find the x and y components of the vector d 1 y N d 1 = 24m d 1y W 30º d 1x E x S

124 Slide 124 / m, 37º east of north Find the x and y components of the vector d 2 y N d 2y 37º d 2 = 28m W E d 2x x S

125 Slide 125 / m, 50º west of south Find the x and y components of the vector d 3 y N W d 3x E d 3 =20m 50º d 3y x S

126 Slide 126 / 246 Using the results from each set of vector components, we can create a table to find the resultant vector components and the magnitude and direction: x (m) y (m) d d d Ʃ d = (d x 2 + d y2 ) d = ( ) d = 31 m tan θ = d y / d x tan θ = 21.5 / 22.4 tan θ = 0.96 θ = tan -1 (0.96) θ = 43.8º

127 Slide 127 / 246 Graphically determine the magnitude and direction of the resultant of the following three vector displacements: d m, 30º north of east d m, 37º north of east d m, 45 o north of east x-component y-component d1 d2 d3 Σ

128 Slide 128 / Graphically determine the magnitude and direction of the resultant of the following three vector displacements: m, 30º north of east m, 37º north of east m, 45 o north of east Magnitude =? 59.7 m / njc.tl/ 58

129 Slide 129 / Graphically determine the magnitude and direction of the resultant of the following three vector displacements: m, 30º north of east m, 37º north of east m, 45 o north of east Direction =? 38.9 o / njc.tl/ 58

130 Slide 130 / Which of the following is an accurate statement? A B C D A vector cannot have a magnitude of zero if one of its components is not zero. The magnitude of a vector can be equal to less than the magnitude of one of its components. If the magnitude of vector A is less than the magnitude of vector B, then the x-component of A must be less than the x-component of B. The magnitude of a vector can be either positive or negative. / njc.tl/ 59

131 Slide 131 / 246 Projectile Motion Return to Table of Contents

132 Slide 132 / 246 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity. Its path is a parabola. v y v x v a = g Projectile Motion Vertical fall

133 Slide 133 / 246 Projectile Motion Projectile motion can be understood by analyzing the vertical and horizontal motion separately. a = g v x The speed in the x-direction is constant. v y v Projectile Motion The speed in the y-direction is changing. Vertical fall

134 Slide 134 / 246 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land? 4.5 m/s 7.5 m x x 0 = 0 v x0 = 4.5 m/s a x = 0 x =? y y 0 = 7.5 m v y0 = 0 a y = -9.8 m/s 2 y = 0 x =?

135 Slide 135 / 246 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land? x y 4.5 m/s x 0 = 0 v x0 = 4.5 m/s a x = 0 x =? y 0 = 7.5 m v y0 = 0 a y = -9.8 m/s 2 y = m y = y 0 + v y0 t + ½ a y t 2 0 = y 0 + ½ a y t 2 x =? t = (-2y 0 /a y ) t = (-2*7.5m/s / -9.8m/s 2 ) t = 1.24 s

136 Slide 136 / 246 A mountain lion leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far from the base of the rock will he land? x y 4.5 m/s 7.5 m x =? x 0 = 0 v x0 = 4.5 m/s a x = 0 x =? t = 1.24 s x = x 0 +v x0 t + ½at 2 x = v x0 t x = (4.5m/s)(1.24s) x = 5.58 m y 0 = 7.5 m v y0 = 0 a y = -9.8 m/s 2 y = 0 y = y 0 + v y0 t + ½ at 2 0 = y 0 + ½ at 2 t = (-2y 0 /a y ) t = (-2*7.5m/s / -9.8m/s 2 ) t = 1.24 s

137 Slide 137 / 246 A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land.

138 Slide 138 / 246 A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land. x x 0 = 0 v x0 = 20 m/s a = 0 x =? y y 0 = 15 m v y0 = 0 a = g = 9.8 m/s 2 y = 0

139 Slide 139 / 246 A cannon ball is shot from a cannon at a height of 15 m with a velocity of 20 m/s. How far away will the cannon ball land. x x 0 = 0 v x0 = 20 m/s a = 0 x =? y y 0 = 15 m v y0 = 0 a = g = -9.8 m/s 2 y = 0 t = 1.75 s y = y 0 + v y0 t + ½ at 2 x = x 0 +v x0 t + ½at 2 y 0 + ½ at 2 x = v x0 t t = (-2y 0 /a) x = (20m/s)(1.75s) t = (-2*15m/-9.8m/s x = 34 m 2 ) t = 1.75 s

140 Slide 140 / 246 Projectile Motion If an object is launched at an angle with the horizontal, the analysis is similar except that the initial velocity has a vertical component. v y v v x = v v x v y v v x v y v v x v x v y v

141 Slide 141 / 246 Projectile Motion - Solving Problems 1. Read the problem carefully. 2. Draw a diagram. 3. Choose an origin and coordinate system. 4. Choose a time interval which is the same for both directions. 5. List known and unknown quantities in both directions. Remember v x never changes and v y at the top is zero. 6. Plan how you will proceed. Choose your equations.

142 Slide 142 / 246 v y v v x = v v x v y v v x v y v v x v x v y v Projectile motion can be described by two kinematics equations: horizontal component: vertical component: x = x 0 + v 0x t + ½a x t 2 y = y 0 + v 0y t + ½a y t 2 v x = v 0x + a x t v y = v 0y + a y t

143 Slide 143 / 246 v y v v x = v v x v y v v x v y v v x v x We can simplify these equations for each problem. For example: v y v x - direction y - direction x 0 = 0 y 0 = 0 v 0x = v 0 cosθ v 0y =v 0 sinθ a x =0 a y = -g x = v 0 cosθt v x = v 0 cosθ y = v 0 sinθt - 1 / 2 gt 2 v y = v 0 sinθ - gt

144 Slide 144 / 246 v y v v x = v v x v y v v x v y v v x v x v y v Flying time: y = v 0 sinθt - ½gt 2 and y = 0 0 =v 0 sinθt - ½gt 2 Now we solve for time: t = (2v 0 sinθ)/g

145 Slide 145 / 246 v y v v x = v v x v y v v x v y v v x v x v y v Horizontal range: In order to find the horizontal range we need to substitute flying time in x(t). x = v 0 cosθt x = v 0 cosθ2v 0 sinθ/g x = 2 v 02 cosθsinθ/g or x = v 02 sin2θ/g

146 Slide 146 / 246 v y v v x = v v x v y v v x v y v v x v x v y v Maximum height: In order to find the maximum height we need to substitute a half of flying time in y(t). y = v 0 sinθt - ½gt 2 y = v 0 sinθ(2v 0 sinθ/g/2) - ½g(2v 0 sinθ/g/2) 2 y = v 02 sin 2 θ/g - v 02 sin 2 θ/2g y = v 02 sin 2 θ/2g

147 Slide 147 / Ignoring air resistance, the horizontal component of a projectile's velocity A B C D is zero remains constant continuously increases continuously decreases / njc.tl/ 5a

148 Slide 148 / A ball is thrown with a velocity of 20 m/s at an angle of 60 above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory? A B C D 10 m/s 17 m/s 20 m/s zero / njc.tl/ 5b

149 Slide 149 / Ignoring air resistance, the horizontal component of a projectile's acceleration A B C D is zero remains a non-zero constant continuously increases continuously decreases / njc.tl/ 5c

150 Slide 150 / At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range? A 0 B 30 C 45 D 60 / njc.tl/ 5d

151 Slide 151 / An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30, 40, 60, and 80. Which two throws cause the javelin to land the same distance away? A 30 and 80 B 30 and 70 C 40 and 80 D 30 and 60 / njc.tl/ 5e

152 Slide 152 / You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be A B C D 1.4 times as much half as much twice as much four times as much / njc.tl/ 5f

153 Slide 153 / A ball is thrown at an original speed of 8.0 m/s at an angle of 35 above the horizontal. What is the speed of the ball when it returns to the same horizontal level? A B C D 4.0 m/s 8.0 m/s 16.0 m/s 9.8 m/s / njc.tl/ 5g

154 Slide 154 / When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed? A B C D It is zero It is less than its initial speed It is equal to its initial speed It is greater than its initial speed / njc.tl/ 5h

155 Slide 155 / A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below? A B C D It is impossible to tell from the information given the stone the ball Neither, since both are traveling at the same speed / njc.tl/ 5i

156 Slide 156 / A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground? A B C D 100 m 170 m 180 m 210 m / njc.tl/ 5j

157 Slide 157 / 246 A projectile is fired with an initial speed of 30 m/s at an angle of 30 o above the horizontal. a. Determine the total time in the air. b. Determine the maximum height reached by the projectile. c. Determine the maximum horizontal distance covered by the projectile. d. Determine the velocity of the projectile 2 s after firing.

158 Slide 158 / 246 A projectile is fired with an initial speed of 30 m/s at an angle of 30 o above the horizontal. a. Determine the total time in the air. v 0 = 30 m/s θ = 30 0 v 0y = v sin (θ) v 0y = 30 sin(30) v 0y = 15 m/s v y-top = v y0 + a y t top t top = (v top - v y0 ) a y x v x = v cosθ y v 0y = v sinθ v y-top = 0 m/s a y = -9.8 m/s 2 t =? t top = (0m/s - 15m/s) 9.8m/s 2 t top = 1.5 s t total = 2 t top t total = 3 s

159 Slide 159 / 246 A projectile is fired with an initial speed of 30 m/s at an angle of 30 o above the horizontal. b. Determine the maximum height reached by the projectile. v 0 = 30 m/s θ = 30 0 t total = 3 s t top = 1.5s (from previous) y = y 0 + v 0y t + 1/2 a y t 2 x v x = v cosθ a x = 0 y v 0y = 15 m/s v y-top = 0 m/s a y = -9.8 m/s 2 y = 0 + (15)(1.5) + 1/2(-9.8)(1.5) 2 y = 22.5m - 11.m y =11.5 m

160 Slide 160 / 246 A projectile is fired with an initial speed of 30 m/s at an angle of 30 o above the horizontal. c. Determine the maximum horizontal distance covered by the projectile. v 0 = 30 m/s θ = 30 0 t total = 3 s x v x = v cosθ a x = 0 y v 0y = 15 m/s v y-top = 0 m/s a y = -9.8 m/s 2 y = 11.5 m v x = v cos θ v x = 30 cos (30 0 ) v x = 26 m/s x = x 0 + v 0x t + 1/2 a x t 2 x = 0 + v 0x t + 0 x = v 0x t x = (26 m/s)(3 s) x = 78 m

161 Slide 161 / 246 A projectile is fired with an initial speed of 30 m/s at an angle of 30 o above the horizontal. d. Determine the velocity of the projectile 2 s after firing. v 0 = 30 m/s θ =30 0 magnitude t total = 3 s v f = (v xf 2 + v yf2 ) direction x v x = 26 m/s a x = 0 y v 0y = 15 m/s v y-top = 0 m/s a y = -9.8 m/s 2 y = 11.5 m v xf = v x0 = 26 m/s v yf = v y0 + a t tan (θ) = (v y /v x ) tan (θ) = 4.6/26 θ = tan -1 (4.6/26) v yf = 15 + (-9.8 m/s 2 )(2s) θ = 10 0 below the horizontal v yf = -4.6 m/s v f = [ (26) 2 +(-4.6) 2 ] v f = ( ) v f = 26.4 m/s

162 Slide 162 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. a. Determine the maximum height reached by the projectile. b. Determine the total time in the air. c. Determine the maximum horizontal distance covered by the projectile. d. Determine the velocity of the projectile just before it hits the bottom of the cliff. / njc.tl/ 5x

163 Slide 163 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. v 0 =30 m/s θ = 45 0 Δy y 0 = 200 m Δx x v x = v cosθ a x = 0 y v 0y = v sin θ v y-top = 0 m/s a y = -9.8 m/s 2 y 0 = 200 m y f = 0m

164 Slide 164 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. a. Determine the maximum height reached by the projectile. x v x = v cosθ a x = 0 y v 0y = v sin θ v y-top = 0 m/s a y = -9.8 m/s 2 y 0 = 200 m y f = 0m v y 2 = v 0y a (y - y 0 ) v y 2 - v 0y 2 = 2a (y - y 0 ) v y 2 - v 0y 2 = (y - y 0 ) 2a y = (v y 2 - v 0y2 ) + y 0 2a y = [0 - (30 sin(45 0 ) 2 ]/ y = 22.9m + 200m y = 222.9m

165 Slide 165 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. b. Determine the total time in the air. x v x = v cosθ a x = 0 y v 0y = v sin θ v y-top = 0 m/s a y = -9.8 m/s 2 y 0 = 200 m y f = 0m y top = m t air = t top + t bottom t top : t bottom : v ytop = v 0y + a t top y bot = y top + v y-top t + 1/2 a y t 2 bot t top = (v y-top - v 0y )/a y bot = y top + 1/2 a y t 2 bot t top = (0 - v sinθ) / a t top = [- 30 sin(45)] / t top = 2.16s t bot = [ (y bot - y top ) ] 2a t bottom = [( )/(2*-9.8)] t bottom = 6.75 s t air = 2.16 s s t air = 8.91 s

166 Slide 166 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. c. Determine the maximum horizontal distance covered by the projectile. x v x = v cosθ a x = 0 t air = 5.7 s y v 0y = v sin θ v y-top = 0 m/s a y = -9.8 m/s 2 y 0 = 200 m y f = 0m y top = m x = x 0 + v 0x t + 1/2 a x t 2 x = v 0x t x = v cosθ t x = 30 cos(45) (8.91s) x = m

167 Slide 167 / A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45 o above the horizontal. d. Determine the velocity of the projectile just before it hits the bottom of the cliff. x y magnitude v f = (v xf 2 + v yf2 ) direction tan (θ) = (v y /v x ) v x = v cosθ a x = 0 v 0y = v sin θ v y-top = 0 m/s a y = -9.8 m/s 2 y 0 = 200 m y f = 0m y top = m t air = 5.7 s v xf = v x0 = 21.2 m/s v yf = v y0 + a t v yf = (-9.8 m/s 2 )(8.91) v yf = m/s tan (θ) = -34.7/21.2 tan(θ) = 1.64 θ = tan -1 (1.64) θ = v f = [ (21.2) 2 +(-66.1) 2 ] v f = ( ) v f = 69.4 m/s

168 Slide 168 / 246 General Problems Return to Table of Contents

169 Slide 169 / A bicyclist moves a long a straight line with an initial velocity v0 and slows down. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration A Positive Negative Negative B Positive Positive Negative C Negative Positive Negative D Negative Negative Positive E Negative Negative Negative / njc.tl/ 5k

170 Slide 170 / A projectile is fired at 60 above the horizontal line with an initial velocity v 0. At which of the following angles the projectile will land at the same distance as it is landed in the first trial? A B C D E / njc.tl/ 5l

171 Slide 171 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. a. How far will the motorcycle travel before catching the car? b. How fast will it be going at that time? c. How does that compare to the car s velocity? d. Draw the following graphs for the car: x(t), v(t), a(t). e. Draw the following graphs for the motorcycle: x(t), v(t), a(t). f. Write the equation of motion for the car. g. Write the equation of motion for the motorcycle. / njc.tl/ 5m

172 Slide 172 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. a. How far will the motorcycle travel before catching the car? / njc.tl/ 5m

173 Slide 173 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. b. How fast will it be going at that time?. / njc.tl/ 5m

174 Slide 174 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. c. How does that compare to the car s velocity? / njc.tl/ 5m

175 Slide 175 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. d. Draw the following graphs for the car: x(t), v(t), a(t). / njc.tl/ 5m

176 Slide 176 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. e. Draw the following graphs for the motorcycle: x(t), v(t), a(t). / njc.tl/ 5m

177 Slide 177 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. f. Write the equation of motion for the car. / njc.tl/ 5m

178 Slide 178 / A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a constant acceleration of 2.4 m/s 2. g. Write the equation of motion for the motorcycle. / njc.tl/ 5m

179 Slide 179 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. a. Indicate every time interval for which speed (magnitude of the velocity) of the cart is decreasing. b. Indicate every time at which the cart is at rest. c. Determine the horizontal position x of the cart at t = 4 s if the cart is located at x 0 = when t 0 = 0. d. Determine the traveled distance of the cart over 10 s from the beginning. e. Determine the average speed of the cart for this time interval. f. Find the acceleration of the cart during time: 0 s -4 s, 4 s - 8 s, 8 - s 10 s, 10 s - 14 s, 14 s - 16 s, 16 s - 20 s. g. On the axes below, sketch the acceleration graph for the motion of the cart from t = 0 s to t = 20 s. / njc.tl/ 5n

180 Slide 180 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. a. Indicate every time interval for which speed (magnitude of the velocity) of the cart is decreasing. t = 0 s to t = 20 s. / njc.tl/ 5n

181 Slide 181 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. b. Indicate every time at which the cart is at rest. / njc.tl/ 5n

182 Slide 182 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. c. Determine the horizontal position x of the cart at t = 4 s if the cart is located at x 0 = when t 0 = 0. / njc.tl/ 5n

183 Slide 183 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. d. Determine the traveled distance of the cart over 10 s from the beginning. / njc.tl/ 5n

184 Slide 184 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. e. Determine the average speed of the cart for this time interval. / njc.tl/ 5n

185 Slide 185 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. f. Find the acceleration of the cart during time: 0 s -4 s, 4 s - 8 s, 8 - s 10 s, 10 s - 14 s, 14 s - 16 s, 16 s - 20 s. / njc.tl/ 5n

186 Slide 186 / A lab cart moves a long a straight horizontal track. The graph describes the relationship between the velocity and time of the cart. g. On the axes below, sketch the acceleration graph for the motion of the cart from t = 0 s to t = 20 s. / njc.tl/ 5n

187 Slide 187 / Find the magnitude and the direction of vector C for the following cases. a. A = 10 N at 0 o, B = 20 N at 0 o, C = A + B b. A = 10 N at 0 o, B = 20 N at180 o, C = A + B c. A = 10 N at 180 o, B = 20 N at 180 o, C = A + B d. A = 10 N at 0 o, B = 20 N at 90 o, C = A + B e. A = 10 N at 90 o, B = 20 N at 0 o, C = A + B / njc.tl/ 5o

188 Slide 188 / Find the magnitude and the direction of vector C for the following cases. a. A = 10 N at 0 o, B = 20 N at 0 o, C = A + B / njc.tl/ 5o

189 Slide 189 / Find the magnitude and the direction of vector C for the following cases. b. A = 10 N at 0 o, B = 20 N at180 o, C = A + B / njc.tl/ 5o

190 Slide 190 / Find the magnitude and the direction of vector C for the following cases. c. A = 10 N at 180 o, B = 20 N at 180 o, C = A + B / njc.tl/ 5o

191 Slide 191 / Find the magnitude and the direction of vector C for the following cases. d. A = 10 N at 0 o, B = 20 N at 90 o, C = A + B / njc.tl/ 5o

192 Slide 192 / Find the magnitude and the direction of vector C for the following cases. e. A = 10 N at 90 o, B = 20 N at 0 o, C = A + B / njc.tl/ 5o

193 Slide 193 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. a. D = 10 N at 37 o Find Dx and Dy. b. E = 20 N at 25 o. Find Ex and Ey. c. Find Gx = Dx + Ex d. Find Gy = Dy + Ey e. Find the magnitude of G from its components f. Find the direction of G. / njc.tl/ 5p

194 Slide 194 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. a. D = 10 N at 37 o Find Dx and Dy. / njc.tl/ 5p

195 Slide 195 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. b. E = 20 N at 25 o. Find Ex and Ey. / njc.tl/ 5p

196 Slide 196 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. c. Find Gx = Dx + Ex / njc.tl/ 5p

197 Slide 197 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. d. Find Gy = Dy + Ey / njc.tl/ 5p

198 Slide 198 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. e. Find the magnitude of G from its components / njc.tl/ 5p

199 Slide 199 / Find the magnitude and the direction of vector G as a sum of two vectors D and E by going through the following steps. f. Find the direction of G. / njc.tl/ 5p

200 Slide 200 / Find the magnitude and the direction of vector C for the following cases. a. A = 40 N at 0 o, B = 10 N at 0 o, C = A + B b. A = 40 N at 0 o, B = 10 N at 180 o, C = A + B c. A = 40 N at 180 o, B = 10 N at 180 o, C = A + B d. A = 40 N at 0 o, B = 10 N at 90 o, C = A + B e. A = 40 N at 90 o, B = 10 N at 0 o, C = A + B / njc.tl/ 5q