Answer:
(a) D(t) = 250t -500 miles
(b) Controller has 2 hours, but including time for pilots to divert course or altitude.
Step-by-step explanation:
Given:
two planes at same altitude heading in a collision course.
Plane A at 400 miles from collision point at 200 mph
Plane B at 300 miles from collision point at 150 mph.
Theoretical collision happens in
t = 400/200 = 300/150 = 2 hours
Distance ya of plane A from collision point as a function of time in hours
ya(t) = 400 -200t
Distance yb of plane B from collision point as a function of time in hours
yb(t) = 300-150t
(a) Distance between two planes,
Since the two planes are on courses perpendicular to each other, will need using pythagorean theorem
D(t) = sqrt(ya(t)^2+yb(t)^2)
= sqrt((400-200t)^2+(300-150t)^2)
= 250(t-2)
D(t) = 250t -500 miles
b. time available
Time until D(t) = 0
solve D(t) = 0
D(t) = 0
250(t-2) = 0
t = 2 (two hours)
Answer:
a) -500 mph
b) 1/2 h
Step-by-step explanation:
a)[tex]\frac{ (150(-300))+(200(-400))}{\sqrt{150x^{2} +200^{2} } }[/tex]
b) [tex]\frac{\sqrt{150^{2}+200^{2} } }{500}[/tex]
