The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 55 miles west to another airport to find that conditions there are too icy for him to land. He flies 25 miles at 15 degrees east of south and is finally able to land at the third airport. (a) how far and in what direction must he fly the next day to go directly to his original destination? (b) how many extra miles beyond his original flight plan has he flown?
Physics - drwls, Monday, September 13, 2010 at 1:38am
(a) Add the vectors of his 55 mile and 25 miles flights. The reverse of that vector is the direction and distance he must fly the next day.
(b) Add the magnitudes of the three flight vectors in the previous question.
Physics - Henry, Monday, September 13, 2010 at 2:41am
15 deg. E of S = 285 deg CCW from 0 deg
X = HOR = 25cos(285) - 55 = -48.5,
Y = VER = 25sin(285) = -24.15,
tanA = Y/X = -24.15/-48.5 = 0.4979,
A = 26.47 deg. = 206.47 deg.(3rd quad.)
d = Y/sinA = -24.15/sin(206.5)=54.18mi
d = 54.18 mi @ 206.47 deg.
a. 54.18 mi N @ 63.5 deg W of South
Location: 3rd QUADRANT.
B. D = 2 * 54.18 = 108.36 MILES