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. T

permaneceerc

permaneceerc

Answered question

2021-01-15

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 willthe pilot travel in the flight? Assume there is no wind during either flight.

Answer & Explanation

pattererX

pattererX

Skilled2021-01-16Added 95 answers

To add vectors, draw each one as the hypoteneuse of atriangle. Fill in the horiz and vert 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 leftof vertical of length 350. Fill in the vert and horiz legs of thistriangle. Then the horiz part is 350 sin30 and the vert part is 350 cos30
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 and 350cos30)
Draw a new triangle, with the north and west parts as the legs. The hypoteneuse of this triangle is the distance from A to B. You can use the pythag theorem to find the length of the hypot, and you can use your favorite trig function to find the angles.
Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-07Added 2605 answers

Step 1

Net displacement vector

D=D1+2+D3

=[(300i^)+(350)(sin30i^+cos30j^)+(150j^)] km

=[125i^+453.1j^] km

=[125i^+453j^]km

Step 2

(a) direction of the displacement

θnet=tan1(DyDx)

=tan1(453.1 km125 km)

=74.577

74.6

b)magnitude of the displacement

D=Dx2+Dy2

=(125 km)2+(453.1 km)2

=470 km

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