Jack and Jill are out walking around in a field. At noon, Jack is located 800 feet west of Jill. Jack walks east at the rate of 5 feet/sec. while Jill walks north at 4 feet/sec. How fast is the distance between them changing at 12:02 p.m.?

Answer 1

At a time of t seconds after noon,
Jack will have walked 5t ft, while Jill will have walked 4t ft
let the distance between them be d ft
I see a right angled triangle with a base of
800-5t, a height of 4 t and a hypotenuse of d

d^2 = (800-5t)^2 + (4t)^2
2d dd/dt = 2(800-5t)(-5) + 2(4t)(4)

at 12:02 , t = 2
d^2 = 624164
d = appr 790.04

dd/dt = (2(790)(-5) + 64)/(2(790.04))
= -4.959..

the distance is DECREASING at appr 4.96 ft/second at that moment
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