Two cars, one going due east at 25 m / sec and the second going due south at 50/3 m / sec are traveling toward the intersection of the two roads they are driving on. at what rate are the two cars approaching each other at the instant when the first car is 200 m and the second car is 150 m from the intersection?

30 m/s

Step-by-step explanation:

Let's say the distance from the first car to the intersection is x, and the distance from the second car to the intersection is y.

The distance between the cars can be found with Pythagorean theorem:

d² = x² + y²

Taking derivative with respect to time:

2d dd/dt = 2x dx/dt + 2y dy/dt

d dd/dt = x dx/dt + y dy/dt

We know that x = 200, dx/dt = -25, y = 150, and dy/dt = -50/3.

To find dd/dt, we still need to find d.

d² = x² + y²

d² = (200)² + (150)²

d = 250

Plugging everything in:

250 dd/dt = (200) (-25) + (150) (-50/3)

dd/dt = -30

The cars are approaching each other at a rate of 30 m/s at that instant.

the solution is 12x-6y