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anonymous
 4 years ago
Plane Ride
You are traveling on an airplane. The velocity of the plane with respect to the air is 110 m/s due east. The velocity of the air with respect to the ground is 43 m/s at an angle of 30° west of due north.
1)What is the speed of the plane with respect to the ground?
2)What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east).° East of due North
3)How far east will the plane travel in 1 hour?
anonymous
 4 years ago
Plane Ride You are traveling on an airplane. The velocity of the plane with respect to the air is 110 m/s due east. The velocity of the air with respect to the ground is 43 m/s at an angle of 30° west of due north. 1)What is the speed of the plane with respect to the ground? 2)What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east).° East of due North 3)How far east will the plane travel in 1 hour?

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327384470550:dwthis is a classic vector addition problem. The first thing I do here is break these into components. We are given a specific grid to use in this case, so we will impose that on our graph and work from that...

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327384677752:dwwe are told that the up direction is zero degrees, and right (east) is positive, so we have\[\overrightarrow{V_p}=V_p\cos\theta\overrightarrow{i}+V_p\sin\theta\overrightarrow{j}=110\cos0^{\circ}\overrightarrow{i}+110\sin0^{\circ}\overrightarrow{j}\]for the planes velocity vector, and\[\overrightarrow{V_a}=V_a\cos\theta\overrightarrow{i}+V_a\sin\theta\overrightarrow{j}=43\cos(30^{\circ})\overrightarrow{i}+43\sin(30^{\circ})\overrightarrow{j}\]The total velocity will be given by the vector sum of these two velocities\[\overrightarrow{V}=\overrightarrow{V_p}+\overrightarrow{V_a}\]The magnitude of this vector is its length, which can be found using the pythagorean theorem with the components\[\overrightarrow{V}^2=i^2+j^2\]so that will be your final speed. The angle can be found by the inverse tangent of its components\[\theta=\tan^{1}(\frac i j)\]so that will be the heading. The distance the plane travels can be found by multiplying the final vectors horizontal speed component (in this case the jcomponent) by time. Just make sure to represent an hour as 3600 seconds to keep units consistent.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1327385738234:dwhere's a little sketch to remind us how the base components i and j go in this case
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