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Vectors and 2 Dimensional Motion

Vectors and 2 Dimensional Motion. There are two kinds of quantities … Vectors are quantities that have both magnitude and direction, such as displacement velocity acceleration Scalars are quantities that have magnitude only , such as position speed time mass. Books usually

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Vectors and 2 Dimensional Motion

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  1. Vectors and2 Dimensional Motion

  2. There are two kinds of quantities… • Vectors are quantities that have both magnitude and direction, such as • displacement • velocity • acceleration • Scalars are quantities that have magnitude only, such as • position • speed • time • mass

  3. Books usually write vector names in bold. You would write the vector name with an arrow on top. This is how you notate a vector… R • Notating vectors R This is how you draw a vector… R head tail

  4. Direction of Vectors • Vector direction is the direction of the arrow, given by an angle. • Draw a vector that has an angle that is between 0o and 90o. θ

  5. Quadrant II 90o < θ< 180o Quadrant I 0o< θ< 90o • Vector angle ranges θ θ θ θ Quadrant III 180o < θ< 270o Quadrant IV 270o < θ< 360o

  6. Direction of Vectors • What angle range would this vector have? • What would be the exact angle, and how would you determine it? Between 180o and 270o θ θ or between -270o and -180o

  7. Magnitude of Vectors • The best way to determine the magnitudeof a vector is to measure its length. • The length of the vector is proportional to the magnitude (or size) of the quantity it represents.

  8. B • If vector A represents a displacement of three miles to the north, then what does vector B represent? Vector C? A C

  9. Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity). • Draw several equal vectors. • Equal Vectors

  10. Inverse vectors have the samelength, but opposite direction. • Draw a set of inverse vectors. A • Inverse Vectors - A

  11. The Right Triangle hypotenuse Opposite side θ adjacent side

  12. Pythagorean Theorem hypotenuse Opposite side θ adjacent side hypotenuse2 = opposite2+ adjacent2

  13. SOHCAHTOA • Basic Trigonometry functions hypotenuse Opposite side θ adjacent side sin θ= opposite/hypotenuse cosθ= adjacent/hypotenuse tan θ= opposite/adjacent sin θ= O / H cosθ= A / H tan θ= O / A

  14. SOHCAHTOA • Inverse functions hypotenuse Opposite side θ adjacent side θ= sin-1(opposite/hypotenuse) θ= cos-1(adjacent/hypotenuse) θ= tan-1(opposite/adjacent) θ= sin-1(O / H) θ= cos-1(A / H) θ= tan-1(O / A)

  15. A surveyor stands on a riverbank directly across the river from a tree on the opposite bank. She then walks 100 m downstream, and determines that the angle from her new position to the tree on the opposite bank is 50o. How wide is the river, and how far is she from the tree in her new location? 100m tan θ = opposite/adjacent 50° ?m tan θ = O / A O = A tan θ O = 100m tan (50°) = 119.18m

  16. You are standing at the very top of a tower and notice that in order to see a manhole cover on the ground 50 meters from the base of the tower, you must look down at an angle 75o below the horizontal. If you are 1.80 m tall, how high is the tower? 1.8m 75° 15° ?m tan θ = O / A 75° O = A tan θ 50m O = 75m tan (75°) = 279.90 m -1.8m = 278.10 m

  17. Graphical Addition of Vectors 1) Add vectors A and B graphically by drawing them together in a head to tail arrangement. 2) Draw vector A first, and then draw vector Bsuch that its tail is on the head of vector A. 3) Then draw the sum, or resultant vector, by drawing a vector from the tail of Ato the head of B. 4) Measure the magnitude and direction of the resultant vector.

  18. B • Practice Graphical Addition A B R A + B = R R is called the resultant vector!

  19. The Resultant and the Equilibrant • The sum of two or more vectors is called the resultant vector. • The resultant vector can replace the vectors from which it is derived. • The resultant is completely canceled out by adding it to its inverse, which is called the equilibrant.

  20. B • Practice Graphical Addition A -R R A + B = R The vector - Ris called the equilibrant. If you add R and -R you get a null (or zero) vector.

  21. Graphical Subtraction of Vectors 1) Subtract vectors A and Bgraphically by adding vector A with the inverse of vector B(-B). 2) First draw vector A, then draw –B such that its tail is on the head of vector A. 3) The difference is the vector drawn from the tail of vector A to the head of -B.

  22. B • Practice Graphical Subtraction B C A A - B = C

  23. Vector A points in the +x direction and has a magnitude of 75 m. Vector B has a magnitude of 30 m and has a direction of 30o relative to the x axis. Vector C has a magnitude of 50 m and points in a direction of -60orelative to the x axis.

  24. dy (m) 140 120 a. Find A + B 100 80 60 40 20 R A B 0 20 40 60 80 100 120 140 dx (m)

  25. dy (m) 140 120 a. Find A + B +C 100 80 60 40 A B C 20 R 0 20 40 60 80 100 120 140 dx (m)

  26. dy (m) 140 120 a. Find A - B 100 80 60 40 A 20 R B 0 20 40 60 80 100 120 140 dx (m)

  27. Vectors: x-component • The x-component of a vector is the “shadow” it casts on the x-axis. • cosθ = adjacent ⁄ hypotenuse • cosθ = Ax ⁄ A • Ax = A cosθ A θ x Ax

  28. Vectors: y-component • The y-component of a vector is the “shadow” it casts on the y-axis. • sin θ = opposite ⁄ hypotenuse • sin θ = Ay ⁄ A • Ay = A sin θ x A Ay Ay θ y

  29. Vectors: angle • The angle a vector makes with the x-axis can be determined by the components. • It is calculated by the inverse tangent function • θ = tan-1 (Ry/Rx) y Ry θ x Rx

  30. Vectors: magnitude • The magnitude of a vector can be determined by the components. • It is calculated using the Pythagorean Theorem. • R2 = Rx2 + Ry2 y R Ry x Rx

  31. You are driving up a long inclined road. After 1.5 miles you notice that signs along the roadside indicate that your elevation has increased by 520 feet. a) What is the angle of the road above the horizontal? b) How far do you have to drive to gain an additional 150 feet of elevation? 1.5 miles ∙ 5280 ft / 1mile = 7920 ft 1.5 miles θ = sin-1 (O / H) 520 ft θ = sin-1 (520 ft / 7920ft) = 3.76° Θ = ? sin θ = O / H H = O / sin θ H = (520 ft + 150 ft) / sin (3.76°) = 10204.62 ft

  32. Find the x- and y-components of the following vectors a) R = 175 meters @ 95º O = H sin θ O = 175 m sin (95°) = 174.33 m 175 m ? m O = H sin θ O = 175 m sin (85°) = 174.33 m 95° 85° ? m A = H cos (θ) A = 175 m cos (95°) = -15.25 m A = H cos (θ) A = 175 m cos (85°) = 15.25 m

  33. Find the x- and y-components of the following vectors b) v = 25 m/s @ -78o O = H sin θ 78° ? m ? m O = 25 m / s ∙ sin (78°) = 24.45 m / s 25 m / s O = H sin θ O = 25 m / s ∙ sin (-78°) = -24.45 m / s A = H cos (θ) A = 25 m / s ∙ cos (78°) = 5.2 m / s A = H cos (θ) A = 25 m / s ∙ cos (-78°) = 5.2 m /s

  34. Find the x- and y-components of the following vectors c) a = 2.23 m/s2 @ 150º O = H sin θ O = 2.23 m / s2 ∙ sin (30°) = 1.12 m / s2 2.23 m / s2 O = H sin θ ? m O = 2.23 m / s2 ∙ sin 150°) = 1.12 m / s2 150° 30° ? m A = H cos (θ) A = 2.23 m / s2 ∙ cos (30°) = 1.93 m / s2 A = H cos (θ) A = 2.23 m / s2 ∙ cos 150°) = -1.93 m /s2

  35. Component Addition of Vectors 1) Resolve each vector into its x- and y- components. Ax = A cosθ Ay = A sin θ Bx = B cos θ By = B sin θ Cx = C cos θ Cy = C sin θ etc. 2) Add the x-components (Ax, Bx, etc.) together to get Rx and the y-components (Ay, By, etc.) to get Ry. 3)Calculate the magnitude of the resultant with the Pythagorean Theorem (R = √(Rx2 + Ry2). 4) Determine the angle with the equation θ = tan-1(Ry/Rx).

  36. In a daily prowl through the neighborhood, a cat makes a displacement of 120 m due north, followed by a displacement of 72 m due west. Find the displacement required if the cat is to return home. 72 m R2 = x2 + y2 R =√( x2 + y2) R = ? m 120 m Θ = ? R =√( (72 m)2 + (120m)2) R = 139.94 m tan θ = O / A θ = tan-1(O / A) θ = tan-1(72 m / 120 m) = 30.96°

  37. If the cat in the previous problem takes 45 minutes to complete the first displacement and 17 minutes to complete the second displacement, what is the magnitude and direction of its average velocity during this 62-minute period of time? v = d/t = 1.6 m / min = 72 m / 45 min v = d/t = 7.06 m / min = 120 m / 17 min 1.6 m / min R2 = x2 + y2 R =√( x2 + y2) R =√( (1.6 m / min)2 + (7.06 m / min)2) R = ? m 7.06 m / min R = 7.24 m / min Θ = ? R = 0.121 m / s tan θ = O / A θ = tan-1(O / A) θ = tan-1((1.6 m /min) / (7.06 m / min)) = 12.77°

  38. If the cat in the previous problem takes 45 minutes to complete the first displacement and 17 minutes to complete the second displacement, what is the magnitude and direction of its average velocity during this 62-minute period of time? 72 m R = 139.94 m v= 139.94 m / 62 min R = ? m 120 m v= 2.26 m / min Θ = ? this is incorrect the velocity is different for each direction!!!!!

  39. Relative Motion

  40. Relative Motion • Relative motion problems are difficult to do unless one applies vector addition concepts. • Define a vector for a swimmer’s velocity relative to the water, and another vector for the velocity of the water relative to the ground. Adding those two vectors will give you the velocity of the swimmer relative to the ground.

  41. Relative Motion Vw Vs Vw Vt = Vs + Vw

  42. Relative Motion Vs Vw Vw Vt = Vs + Vw

  43. Relative Motion Vs Vw Vw Vt = Vs + Vw

  44. Vt = Vs + Vw • You are paddling a canoe in a river that is flowing at 4.0 mph east. You are capable of paddling at 5.0 mph. a) If you paddle east, what is your velocity relative to the shore? b) If you paddle west, what is your velocity relative to the shore? 4.0 mph 5.0 mph Vt = 5.0 mph + 4.0 mph Vt = 9.0 mph Vt = Vs + Vw 4.0 mph Vt = (-5.0 mph) + 4.0 mph 5.0 mph Vt = -1.0 mph

  45. You are paddling a canoe in a river that is flowing at 4.0 mph east. You are capable of paddling at 5.0 mph. c) You want to paddle straight across the river, from the south to the north. At what angle do you aim your boat relative to the shore? Assume east is 0o. 5.0 mph θ tan θ = O / A 4.0 mph θ = tan-1 (O / A) θ = tan-1 (5.0 mph / 4.0mph) = 51.34°

  46. 80 mph • You are flying a plane with an airspeed of 400 mph. If you are flying in a region with a 80 mph west wind, what must your heading be to fly due north? Tan θ = O / A θ = Tan-1 (O / A) 400 mph θ θ = Tan-1 (80 mph / 400 mph) = 11.31°

  47. 2-Dimensional Motion • Definition: motion that occurs with both x and y components. • Examples: • Playing pool (billiards). • Throwing a ball to another person. • Each dimension of the motion can obey different equations of motion.

  48. Drawing Vectors • You’ve known since that displacement vectors can be composed of horizontal and vertical components • Etch-A-Sketch physics • http://www.etchy.org/

  49. Keys to Solving 2-D Problems 1. Resolve all vectors into components • x-component • Y-component 2. Work the problem as two one-dimensional problems. • Each dimension can obey different equations of motion. 3. Re-combine the results for the two components at the end of the problem.

  50. You run in a straight line at a speed of 5.0 m/s in a direction that is 40° south of west. a) How far west have you traveled in 2.5 minutes? v = ? m/s T = 2.5 min = 150 s θ = 40° cosθ = A / H 5 m/s A = H cosθ A = 5 m/s cos (40°) = 3.83 m/s v = d/t d = v ∙ t d = 3.83 m/s ∙ 150s = 574 m

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