related rates

Question

A person is 8 meters away from a road and there is a car that is initially 800 meters away approaching the person at a speed of 45 m/sec. At what rate is the distance between the person and the car changing (a) 5 seconds after the start, (b) when the car is directly in front of the person and (c) 10 seconds after the car has passed the person. See the (probably bad) sketch below to help visualize the problem.

Answer 1

Distance between the car and the man = s = sqrt [(8²+(800-x)²], where x is the distance measured from the starting point.

Differentiate the equation with respect to t, we get

ds/dt = 1/2 (-1600+2x)/sqrt[64+(800 - x)² ] dx/dt

ds/dt is the changing rate between the person and the car and dx/dt is the speed of the car.

(a)  5 seconds after the start, x = 45*5 = 225

      ds/dt = (x - 800)/sqrt [65 + (800-x)²] *45 = -45 m/s  (negative sign means decreasing rate)

(b)  Just in front x = 800, ds/dt = 0

(c)  10 seconds after passing the person x = 450, also s = sqrt [8²+x²]

      ds/dt = 1/(2 sqrt  [8²+x²])(2x) dx/dt

      ds/dt = 45 m/s