Two cars are traveling along perpendicular roads, car A at 40 mph, car B at 60 mph. At noon, when car A reaches the intersection, car B is 90 miles away and moving toward it. Find the rate at which the distance between the cars is changing at 1 P. M..

-4

Explanation:

Parameters given:

Speed of car A = 40 mph

Speed of car B = 60 mph

Initial position of car B relative to A = 90 miles

The attached diagram explains better.

For the sake of this solution, we assume that car A is moving on the y axis and car B is moving on the x axis.

First, calculate the position of both cars after 1 hour:

Car A:

y = 40 * 1 = 40 miles

Car B:

(60 * 1) = 60 miles.

Since car B is moving towards A and it was initially 90 miles, the position of B will be (90 - 60) miles

x = 30 miles

We can find the relative distance of both cars (s in the diagram) using Pythagoras theorem:

s² = x² + y²

s² = 30² + 40² = 2500

s = 50 miles

To find the rate at which this distance is changing with time, we differentiate the Pythagoras equation with respect to time:

s² = x² + y²

=> 2s*ds/dt = 2x*dx/dt + 2y*dy/dt

dx/dt = -60 mph (because car B is moving towards car A, hence it is approaching the -ve x axis)

dy/dt = 40 mph (it moves upward on the +ve y axis)

=> 2*50*ds/dt = (2*30*-60) + (2*40*40)

100*ds/dt = -3600 + 3200

100*ds/dt = -400

=> ds/dt = -400/100

ds/dt = -4

The negative sign indicates that the distance between both cars (s) is decreasing.

it may be wrong but i think its a

answer: option c.

explanation: