## 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?

1. TuringTest

|dw:1327384470550:dw|this 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...

2. TuringTest

|dw:1327384677752:dw|we 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 j-component) by time. Just make sure to represent an hour as 3600 seconds to keep units consistent.

3. TuringTest

|dw:1327385738234:dw|here's a little sketch to remind us how the base components i and j go in this case