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?

To solve this problem, we can consider the relative motion between the two planes.

Let's break down the problem step by step:

a. How fast is the private plane flying?

To find the speed of the private plane, we need to calculate its velocity components in the east and north directions.

The commercial freight carrier is traveling 4 miles east for every 3 miles north, which gives us a velocity vector of (4, 3). The private plane is traveling 12 miles north for every 5 miles west, which gives us a velocity vector of (-5, 12).

To find the relative velocity of the private plane with respect to the commercial carrier, we can subtract the velocity vector of the commercial carrier from the velocity vector of the private plane:

Relative velocity = (velocity of private plane) - (velocity of commercial carrier)
= (-5, 12) - (4, 3)
= (-9, 9)

The relative velocity vector of (-9, 9) indicates that for every 9 miles the private plane moves north, it will move 9 miles east. This means the private plane is moving at a speed of 9 miles per hour.

Therefore, the private plane is flying at a speed of 9 miles per hour.

b. When will the collision take place if it is not averted?

To determine the time of collision, we can calculate the distance between the two planes and divide it by the relative speed of the private plane.

The distance between the two planes is given as 210 miles east of the commercial carrier.

Since the private plane is moving at a relative velocity of (-9, 9), it means it covers 9 miles in both directions in the same amount of time. So, the total distance traveled by the private plane (east + north) is 9 times the time of collision.

Distance traveled east = 9 times of time of collision = 210 miles

Solving for the time of collision, we have:

9 * time = 210
time = 210 / 9 ≈ 23.33 hours

Therefore, the collision will take place after approximately 23.33 hours if it is not averted.

Please note that this solution assumes both planes continue on their current trajectories without making any turns or adjustments to their paths.

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]
= 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