Asked by Julian
A nuclear submarine leaves its base and travels at 23.5 mph. For 2.00 hrs it travels along a course 32.1 degrees north of west. It then turns an additional 21.5 degrees north of west and travels for another 1.00 hrs. How far from its base is it?
Answers
Answered by
Henry
d1 = 23.5mi/h * 2h = 47 Mi[180o-32.1o] =
47Mi[147.9o]
d2 = 23.5mi/h * 1h = 23.5mi[180-53.6] =
23.5mi[126.4o]
d1 + d2 = 47mi[147.9] + 23.5mi[126.4o]
X = 47*cos147.9 + 23.5*cos126.4=-53.8 mi
Y = 47*sin147.9 + 23.5*sin126.4=43.9 mi
D^2=X^2 + Y^2=-53.8^2 + 43.9^2=4821.65
D = 69.4 miles.
47Mi[147.9o]
d2 = 23.5mi/h * 1h = 23.5mi[180-53.6] =
23.5mi[126.4o]
d1 + d2 = 47mi[147.9] + 23.5mi[126.4o]
X = 47*cos147.9 + 23.5*cos126.4=-53.8 mi
Y = 47*sin147.9 + 23.5*sin126.4=43.9 mi
D^2=X^2 + Y^2=-53.8^2 + 43.9^2=4821.65
D = 69.4 miles.
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