Solve this related rates problem:An automobile traveling at a rate of 20 feet per second is approaching an intersection. when the automobile is 100 ft from the intersection a truck traveling at the rate of 40 ft per second crosses the intersection. the automobile and the truck are on roads that are right angles to each other. how fast are the truck and the automobile separating 2 seconds after the truck leaves the intersection

This is a classic related rates problem that can be solved using the Pythagorean theorem. Here are the steps to solve it:

Step 1: Draw a diagram Draw a right triangle with the intersection as the vertex of the right angle. The automobile is moving along one leg of the triangle, the truck along the other, and the hypotenuse represents the distance between the two vehicles.

Step 2: Set up the Pythagorean theorem Let x be the distance the car is from the intersection, y be the distance the truck is from the intersection, and z be the distance between the car and the truck. According to the Pythagorean theorem, we have x² + y² = z².

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

Step 4: Plug in the known values We know that dx/dt = -20 ft/sec (the car is moving towards the intersection), dy/dt = 40 ft/sec (the truck is moving away from the intersection), x = 100 - 20t (the car is 100 ft away and moving at 20 ft/sec), and y = 40t (the truck is moving at 40 ft/sec). We want to find dz/dt when t = 2 sec.

Step 5: Solve for dz/dt First, plug t = 2 into the equations for x and y to find their values at this time. This gives x = 60 ft and y = 80 ft. Then, use the Pythagorean theorem to find z = sqrt(x² + y²) = 100 ft. Finally, plug all these values into the differentiated equation and solve for dz/dt. This gives dz/dt = (260(-20) + 28040) / (2*100) = 28 ft/sec.

So, 2 seconds after the truck leaves the intersection, the truck and the automobile are separating at a rate of 28 feet per second.