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You can think of this as a Pythagorean Theorem problem. The hypotenuse is the unknown. One short leg is the 150km/h minus the east component of the wind. The other short leg is the north component of the wind:
east comp: 95*cos(45) -- so the total leg is 150-95*cos(45)
north comp: 95*sin(45)
The magnitude of the velocity will be:
sqrt((150-95*cos(45))^2 + (95*sin(45))^2)
= 106.64 km/h
Let me know if you have any questions,
I can help you with that as well.
The magnitude is sqrt(20^2 + 20^2) = 28.28 m
The direction is arctan(-20/20) = -45 degrees (45 degrees south of east) (but that is the same as 315 degrees counterclockwise from east).
Hi again! As far as I can tell, the first answer was accepted and I got the bonus. Thanks!
The magnitude is another Pythagorean Triangle:
sqrt(16^2 + 6^2) = 17.08 m/s
The angle is given by the arctangent:
angle = arctangent(y/x)
arctan(6/16) = 20.56 degrees
Could it be a rounding problem? For example, have you tried it with 20.6 degrees or 20.556 degrees?
Hmmm.... that is strange. Is there anyone you can contact to see if there is a bug in the program?