This problem can be solved using the Pythagorean theorem and the concept of related rates in calculus.

Step 1: Set up the problem
Let's denote the car moving south as car S and the car moving west as car W. Let x be the distance car S has traveled after 3 hours and y be the distance car W has traveled after 3 hours. The distance between the two cars is z.

Step 2: Apply the Pythagorean theorem
According to the Pythagorean theorem, we have x^2 + y^2 = z^2.

Step 3: Differentiate both sides with respect to time (t)
We get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).

Step 4: Plug in the known values
We know that dx/dt = 56 mi/h (the speed of car S), dy/dt = 42 mi/h (the speed of car W), x = 56*3 = 168 miles, and y = 42*3 = 126 miles.

Step 5: Solve for dz/dt
We first need to solve for z using the Pythagorean theorem. z = sqrt(x^2 + y^2) = sqrt((168)^2 + (126)^2) = 210 miles.
Then we can solve for dz/dt. dz/dt = (x*dx/dt + y*dy/dt) / z = (168*56 + 126*42) / 210 = 70 mi/h.

So, the distance between the cars is increasing at a rate of 70 mi/h three hours later.

Question

This problem can be solved using the Pythagorean theorem and the concept of related rates in calculus.

Step 1: Set up the problem
Let's denote the car moving south as car S and the car moving west as car W. Let x be the distance car S has traveled after 3 hours and y be the distance car W has traveled after 3 hours. The distance between the two cars is z.

Step 2: Apply the Pythagorean theorem
According to the Pythagorean theorem, we have x^2 + y^2 = z^2.

Step 3: Differentiate both sides with respect to time (t)
We get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).

Step 4: Plug in the known values
We know that dx/dt = 56 mi/h (the speed of car S), dy/dt = 42 mi/h (the speed of car W), x = 56*3 = 168 miles, and y = 42*3 = 126 miles.

Step 5: Solve for dz/dt
We first need to solve for z using the Pythagorean theorem. z = sqrt(x^2 + y^2) = sqrt((168)^2 + (126)^2) = 210 miles. 
Then we can solve for dz/dt. dz/dt = (x*dx/dt + y*dy/dt) / z = (168*56 + 126*42) / 210 = 70 mi/h.

So, the distance between the cars is increasing at a rate of 70 mi/h three hours later.

Knowee AI · Accepted Answer

This problem can be solved using the Pythagorean theorem and the concept of related rates in calculus.

Step 1: Set up the problem
Let's denote the car moving south as car S and the car moving west as car W. Let x be the distance car S has traveled after 3 hours and y be the distance car W has traveled after 3 hours. The distance between the two cars is z.

Step 2: Apply the Pythagorean theorem
According to the Pythagorean theorem, we have x^2 + y^2 = z^2.

Step 3: Differentiate both sides with respect to time (t)
We get 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt).

Step 4: Plug in the known values
We know that dx/dt = 56 mi/h (the speed of car S), dy/dt = 42 mi/h (the speed of car W), x = 56*3 = 168 miles, and y = 42*3 = 126 miles.

Step 5: Solve for dz/dt
We first need to solve for z using the Pythagorean theorem. z = sqrt(x^2 + y^2) = sqrt((168)^2 + (126)^2) = 210 miles. 
Then we can solve for dz/dt. dz/dt = (x*dx/dt + y*dy/dt) / z = (168*56 + 126*42) / 210 = 70 mi/h.

So, the distance between the cars is increasing at a rate of 70 mi/h three hours later.

Two cars start moving from the same point. One travels south at 56 mi/h and the other travels west at 42 mi/h. At what rate (in mi/h) is the distance between the cars increasing three hours later? mi/h

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