1. Plane A is 40 mi south and 100 mi east of Plane B. Plane A is flying 2 miles west for every mile it flies north, while Plane B is flying 3 mi east for every mile it flies south.

a. Where do their paths cross?

b. Which plane must fly farther?

c. What ratio of the speed of Plane B to the speed of Plane A would
produce a midair collision?

Start by drawing a diagram of the situations, such as:

http://img101.imageshack.us/img101/3371/newt.jpg
a. The point where the two paths cross is at (-40,20) relative to plane A. (Check my graphical solution).
b. not sure what exactly is requested.
Either plane A can fly further south, or plane B a little further north, or they can avoid each other by changing altitudes.
c. The deadly ratio can be found by calculating the distances of each plane from the intersection point (-40,20). If the speeds are inversely proportional to the distance, it is likely to have a mid-air collision.

To solve this problem, we can break it down into three parts: finding the point of intersection, determining which plane must fly farther, and finding the ratio of speeds for a midair collision.

a. To find the point of intersection, we need to track the positions of both planes. Plane A is initially 40 miles south and 100 miles east of Plane B. Plane A flies 2 miles west for every mile it flies north, while Plane B flies 3 miles east for every mile it flies south. Let's denote the distance traveled north by Plane A as "x." Thus, the distance traveled west by Plane A is "2x." The distance traveled south by Plane B is also "x," and therefore, the distance traveled east by Plane B is "3x".

To find the point of intersection, we can set up a system of equations:

For Plane A: Distance traveled north and west, respectively, is x and 2x.
For Plane B: Distance traveled south and east, respectively, is x and 3x.

The coordinates of Plane A can be represented as (100 - 2x, 0 + x), and the coordinates of Plane B can be represented as (0 + 3x, 0 - x). To find the point of intersection, we equate the coordinates:

(100 - 2x, x) = (3x, -x)

We can set the corresponding x-values equal to each other and solve for x:

100 - 2x = 3x

100 = 5x

x = 20

Now that we know x = 20, we can substitute it back into either coordinate equation to find the point of intersection. Let's use Plane A's coordinates:

(100 - 2(20), 20) = (60, 20)

Therefore, the paths of the two planes cross at the point (60, 20).

b. To determine which plane must fly farther, we need to compare the distances traveled by each plane. Plane A flies a total distance of 40 miles north (x = 20), and Plane B flies a total distance of 40 miles south (x = 20). Since both planes travel an equal distance vertically, we can compare their horizontal distances:

Plane A travels 2x = 2 * 20 = 40 miles west.
Plane B travels 3x = 3 * 20 = 60 miles east.

Therefore, Plane B must fly farther.

c. To find the ratio of the speed of Plane B to the speed of Plane A that would produce a midair collision, we need to consider their relative speeds. Plane A's speed is determined by the ratio 2 miles west for every mile north (2:1), while Plane B's speed is determined by the ratio 3 miles east for every mile south (3:1).

To maintain a relative collision course, the ratio of their speeds must be equal to the ratio of the distances they have to cover. Plane A has to travel 40 miles north, while Plane B has to travel 40 miles south. So the ratio of their distances is 1:1.

Thus, the ratio of the speed of Plane B to the speed of Plane A that would produce a midair collision is 3:2.