Velocity MCV4U: Calculus & Vectors Recall that speed, a measure - PDF document

g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Velocity MCV4U: Calculus & Vectors Recall that speed, a measure of how fast something is travelling, is a scalar quantity. Velocity is speed with direction, and is a vector quantity. By using vectors to represent velocities, it is possible to solve a variety of problems. Velocities as Vectors J. Garvin J. Garvin — Velocities as Vectors Slide 1/16 Slide 2/16 g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Resultant Velocity Problems Resultant Velocity Problems Example Use the Pythagorean Theorem to find the kayaker’s speed. A kayaker paddles 8 km/h due south across a river that has a � 8 2 + 5 2 | � r | = current flowing 5 km/h due east. What is the resulting √ velocity of the kayaker? = 89 ≈ 9 . 4 km/h Use the following diagram, where � k is the velocity of the kayaker, � c is the velocity of the current, and � r is the resultant Use a trigonometric ratio to find the direction. velocity. � 5 � θ = tan − 1 8 ≈ 32 ◦ Therefore, the resulting velocity is approximately 9.4 km/h S32 ◦ E. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 3/16 Slide 4/16 g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Resultant Velocity Problems Resultant Velocity Problems Example An airplane is travelling on a bearing of 330 ◦ at a constant speed of 150 km/h. A wind blows on a bearing of 85 ◦ at 40 km/h. Determine the speed and direction of the airplane relative to the ground. Use the following diagram, where � AH is the airplane’s AW is the wind’s velocity, and � � velocity, AR is the resultant velocity of the airplane relative to the ground. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 5/16 Slide 6/16

g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Resultant Velocity Problems Resultant Velocity Problems To determine the velocity of the airplane relative to the To determine the direction, use the sine law (or cosine law) ground, we need to determine the magnitude of � AR . to find the measure of ∠ HAR , then add it to the airplane’s From the given bearings, ∠ WAH = 30 ◦ + 85 ◦ = 115 ◦ , and original bearing. ∠ AHR = 180 ◦ − 115 ◦ = 65 ◦ . sin( HAR ) = sin( AHR ) Use the cosine law to determine | � AR | . | � | � HR | AR | sin( HAR ) ≈ sin(65 ◦ ) � | � | � HR | 2 + | � AH | 2 − 2( | � HR | )( | � AR | = AH | ) cos( AHR ) 40 138 � 40 sin(65 ◦ ) � � 40 2 + 150 2 − 2(40)(150) cos(65 ◦ ) ∠ HAR ≈ sin − 1 = 138 ≈ 138 km/h ≈ 15 ◦ Therefore, the actual bearing of the airplane is approximately 330 ◦ + 15 ◦ , or 345 ◦ . Its actual speed is 138 km/h. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 7/16 Slide 8/16 g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Problems Given a Resultant Velocity Problems Given a Resultant Velocity � Example In the diagram, TM represents the resultant trip from eal. Let � Toronto to Montr´ TB be the velocity of the plane on A pilot wishes to fly 508 km from Toronto to Montr´ eal, on a � its new bearing and let TW be the velocity of the wind. bearing of 78 ◦ . The airplane has a top speed of 550 km/h. An 80 km/h wind is blowing on a bearing of 125 ◦ . • In what direction should the pilot fly to reach the destination? • At what speed will the plane be travelling? • How long will the trip take? Since we are given information about the resultant, we need to work backward! Note that ∠ TMB = ∠ MTW = 12 ◦ + 35 ◦ = 47 ◦ . J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 9/16 Slide 10/16 g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Problems Given a Resultant Velocity Problems Given a Resultant Velocity We can use the sine law to find the measure of ∠ BTM . Now that we know the bearing, we can calculate the speed of the airplane using the cosine law. sin( BTM ) = sin( BMT ) ∠ TBM ≈ 180 ◦ − 47 ◦ − 6 . 1 ◦ ≈ 126 . 9 ◦ . | � | � BM | TB | sin( BTM ) ≈ sin(47 ◦ ) � TB | 2 + | � BM | 2 − 2( | � | � | � TB | )( | � TM | = BM | ) cos( TBM ) 80 550 � � 80 sin(47 ◦ ) � 550 2 + 80 2 − 2(550)(80) cos(126 . 9 ◦ ) ∠ BTM ≈ sin − 1 ≈ 550 ≈ 601 km/h ≈ 6 . 1 ◦ Since Montr´ eal is 508 km away, a plane travelling at 601 The pilot must fly at a bearing of approximately km/h will make the trip in 508 / 601 ≈ 0 . 85 hours, or 51 78 ◦ − 6 . 1 ◦ ≈ 71 . 9 ◦ . minutes. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 11/16 Slide 12/16

g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Relative Velocity Relative Velocity Relative velocity is what an observer perceives when s/he Example perceives her/himself to be stationary. A truck is travelling east at 60 km/h, while a car is travelling south at 80 km/h. What is the relative velocity of the truck It is the difference of two velocities. to the car? Relative Velocity When two objects, A and B , have velocities � v A and � v B , the Let � v T be the velocity of the truck and � v C the velocity of the velocity of B relative to A is � v rel = � v B − � v A . car. We want � v T − � v C , or � v T + ( − � v C ). While − � v C has the opposite direction as � v C , the two vectors are still at right angles to each other. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 13/16 Slide 14/16 g e o m e t r i c v e c t o r s g e o m e t r i c v e c t o r s Relative Velocity Questions? Using the Pythagorean Theorem: � 80 2 + 60 2 | � v rel | = = 100 km/h Using the tangent ratio for the direction: � 60 � θ = tan − 1 80 ≈ 37 ◦ Therefore, the velocity of the truck relative to the car is 100 km/h, at a bearing of approximately N37 ◦ E. J. Garvin — Velocities as Vectors J. Garvin — Velocities as Vectors Slide 15/16 Slide 16/16