Third leg.
The crew flies at a speed of 560 mi/h in direction N-20°-E.
The wind has a speed of 35 mi/h and a direction S-10°-E.
We then can draw this as:
We have to add the two vectors to find the actual speed and direction.
We will start by adding the x-coordinate (W-E axis):
and the y-coordinate (S-N axis) is:
Then, the actual speed vector is v3=(197.61, 491.76).
The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].
Then, we have to calculate the displacement in 20 minutes using the actual speed vector.
We can calculate the movement in each of the axis. For the x-axis:
NOTE: 20 minutes represents 1/3 of an hour.
We can do the same with the y-coordinate:
The final position is R3 = (282.53, 331.59).
To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:
The distance can be calculated as if it was a right triangle:
![\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d%5E2%3Dx%5E2%2By%5E2_%7B%7D%20%5C%5C%20d%5E2%3D282.53%5E2%2B331.59%5E2%20%5C%5C%20d%5E2%3D79823.20%2B109951.93%20%5C%5C%20d%5E2%3D189775.13%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B189775.13%7D%20%5C%5C%20d%5Capprox435.63%20%5Cend%7Bgathered%7D)
The angle, from E to N, can be calculated as:
If we want to express it from N to E, we substract the angle from 90°:
Answer: the final location can be represented with the vector (282.53, 331.59).
1) The distance from the origin is 435.63 miles and
2) the direction is N-40°-E.
