Question 814549: A plane is heading due south with an airspeed of 288 mph. A wind from a direction of 58° is blowing at 20 mph. Find the bearing of the plane.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! use vector addition:
---
wind vector w:
direction from = 58 degrees
direction to = 238 degrees
magnitude = 20
---
decompose w into rectangular coordinates:
wy = 20cos(58) = 10.598
wx = 20sin(58) = 16.961
---
airplane vector a:
direction = 180 degrees
magnitude = 288
---
decompose a into rectangular coordinates:
ay = 288cos(0) = 288
ax = 288sin(0) = 0
---
sum:
sy = ay + wy = 288 + 10.598 = 298.598
sx = ax + wx = 0 + 16.961 = 16.961
---
convert the sum vector to polar form:
s = sqrt( sx^2 + sy^2 ) = 299.079 mph
t = arctan( sx / sy ) = 3.251 degrees
---
Answer:
the airplane will fly a bearing of (180 + 3.251) = 183.251 degrees at a ground speed of 299.079 mph
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Convert fractions, decimals, and percents:
https://sooeet.com/math/fraction-decimal-percent.php
---
Calculate and graph the linear regression of any data set:
https://sooeet.com/math/linear-regression.php
|
|
|
