posted by newt on .
A commercial freight carrier is flying at a constant speed of 400 mi/h
and is traveling 4 miles east for every 3 miles north. A private plane is
observed to be 210 miles east of the commercial carrier, traveling 12
miles north for every 5 miles west.
a. If it is known that the two planes are on a collision course, how fast
is the private plane flying?
b. When will the collision take place if it is not averted?
I placed the carrier on the origin so that plane flies along the line y = (3/4)x
When the carrier is at the origin, the private plane is at (210,0) and flies along a line with slope -12/5
Since I know the slope and a point, I found the line of travel for the private plane to be
y = (-12/5)x + 504
The collision occurs at the intersection of these two lines, which I found by solving to be (160,120)
From that Point to the origin is
√(160^2 + 120^2) = 200
Since the carrier was flying at 400 mph, the collision occurred at t = 200/400 hours = 1/2 hour.
The distance from (160,120) to (210,0)
= √[(210-160)^2 + 120^2]
Since the distance = 130 miles and the time = 1/2 hours,
v = distance/time = 130/(1/2) = 260
The small plane went at 260 mph