At what rate is the distance between the two cars changing two hours later?

Well since both cars start at the same point, but one is moving south and the other west then it will form a triangle . You will have to use the Pythagorean theorem (c2= a2+b2)

Car moving South = 60mi/h

Car moving West = 25 mi/h

So .. 60 mi/h x 2 h = 120 miles and 25 mi/h x 2 h = 50 miles

c2 = a2 + b2

c2 = 120^2 + 50^2

c2 = 14,400 + 2,500 14400+ 2,500 = 16900

c2 = the square root of 16900

square root of 16900 = 130

So the cars have moved 130 miles apart in 2 hours

130/2 = 65 mi/h

The rate of the distance between the two cars is changing at 65 mi/h

Hope this helps :)

The rate of change of the distance between the two cars is basically the RELATIVE VELOCITY between those two cars. So, if you can get that, you get the solution.

Now, lets calculate the relative velocity of the west-car with respect to the south-car.

Velocity of west-car, Vw= i. [25 mi/h]

Velocity of south-car, Vs= j. [60 mi/h], here i, j are just two unit vector, there positive direction are in west and south respectively.

Now, the relative velocity of the west-car with respect to the south-car, Vws = Vw - Vs = i. [25 mi/h]- j. [60 mi/h],

It's magnitude is sq. root of (25^2 + 60^2) = 65 mi/h, this is the solution.

As you can see the solution doesn't have any time related term, hence it doesn't matter whether after 2 hrs or 20 hrs, the rate will be same always as long as they maintain their uniform velocity.

After 2 hours = 50 X50 = 2500

120 X 120 = 14400

ADD = 16900

\/16900 = 130 miles

After 3 hours = 75 X 75 = 5625

180 X 180 = 32400

ADD = 38025

\/38025 = 195 miles

After 4 hours = 100 X 100 = 10,000

240 X 240 = 57600

ADD = 67600

\/67600 = 260 miles

ANSWER 65 mph ( 260-195, 195-130, 130-65)