University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

227

Pages:

1
Author:

banenetta0l

Question 1 An airplane starting from airport A flies 300 km east, then 350 km at 30 degrees west of north, and then 150 km north to arrive finally at airport B. The next day, another plane flies directly from A to B in a straight line. In what direction should the pilot travel in this direct flight? How far will the pilot travel in the flight? Assume there is no wind during either flight. Answer 1 To add vectors, draw each one as the hypotenuse of a triangle. Fill in the horizontal and vertical legs of the triangle. Using trig, figure out the lengths of the legs. For the plane from airport A, draw a line 30 deg to the left of vertical of length 350. Fill in the vertical and horizontal legs of this triangle. Then the horizontal part is 350 sin ⁡ 30 350sin30 and the vertical part is 350 cos ⁡ 30 350cos30. Notice that the plane traveled north and west. You now know how much north and how much west. It also traveled 150 km north on the second part of its flight to arrive at airport B. So now you know how much west (just the one leg) and how much north (add the 150 150 and 350 cos ⁡ 30 350cos30). Draw a new triangle, with the north and west parts as the legs. The hypotenuse of this triangle is the distance from A to B. You can use the Pythagorean theorem to find the length of the hypotenuse and use your favorite trig function to find the angles. Answer 2 Step 1: Net displacement vector 𝐷 ⃗ = 𝐷 1 ⃗ + 𝐷 2 ⃗ + 𝐷 3 ⃗ D = D 1 + D 2 + D 3 = [ ( 300 𝑖 ^ ) + ( 350 ) ( − sin ⁡ 3 0 ∘ 𝑖 ^ + cos ⁡ 3 0 ∘ 𝑗 ^ ) + ( 150 𝑗 ^ ) ] km =[(300 i ^ )+(350)(−sin30 ∘ i ^ +cos30 ∘ j ^ )+(150 j ^ )]km = [ 125 𝑖 ^ + 453.1 𝑗 ^ ] km =[125 i ^ +453.1 j ^ ]km Step 2: (a) Direction of the displacement 𝜃 net = tan ⁡ − 1 ( 𝐷 𝑦 𝐷 𝑥 ) θ net =tan −1 ( D x D y ) = tan ⁡ − 1 ( 453.1 km 125 km ) =tan −1 ( 125km 453.1km ) = 74.5 7 ∘ =74.57 ∘ ≈ 74. 6 ∘ ≈74.6 ∘ (b) Magnitude of the displacement 𝐷 = 𝐷 𝑥 2 + 𝐷 𝑦 2 D= D x 2 +D y 2 = ( 125 km ) 2 + ( 453.1 km ) 2 = (125km) 2 +(453.1km) 2 = 470 km =470km