Monday

July 28, 2014

July 28, 2014

Posted by **newt** on Monday, September 28, 2009 at 10:56am.

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?

- Algebra 2 -
**Reiny**, Monday, September 28, 2009 at 1:01pmI 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]

= 130

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

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