SOLUTION: Ship A was sailing due to South at the rate of 6 miles per hour and another ship B was sailing due to East at the rate of 8 miles per hour. At 4 PM ship B crossed the position wher

Question 826008: Ship A was sailing due to South at the rate of 6 miles per hour and another ship B was sailing due to East at the rate of 8 miles per hour. At 4 PM ship B crossed the position where hip A was 2 hours before.
1) At what rate were they approaching or separating at 3 PM ?
2) At what rate were they changing the distance between them at 5 PM ?
3) When was the distance between them not changing ?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Ship A was sailing due to South at the rate of 6 miles per hour and another ship B was sailing due to East at the rate of 8 miles per hour. At 4 PM ship B crossed the position where hip A was 2 hours before.
Use 2 PM as the starting point.
Ship A is 12 miles North of it, and ship B is at the point.
------
Find the distance between the ships as a function of time (in hours).
t = 0 at the starting point.
---
Ship A's distance from the SP is 12 - 6t miles
Ship B's distance from the SP is 8t miles
The distance between them is the hypotenuse of a right triangle.

Find the relative speed, ds/dt
ds/dt =
s'(t) =
s'(t) =
============================================
1) At what rate were they approaching or separating at 3 PM ?
3 PM is t = 1
s'(1) =
= 14/5 = 2.8 mi/hr
----------------------------------------------
2) At what rate were they changing the distance between them at 5 PM ?
5 PM --> t = 3
s'(3) =
=
=
=~ 9.216 mi/hr
------------------------------------------
3) When was the distance between them not changing ?
s'(t) = = 0
50t-36 = 0
t = 0.72 hrs past 2 PM
t = 2:43:12 PM
-------------------------
The speeds in parts 1 & 2 are both separating, since they're past the time in part 3.
===========================================
It appears to be a simple problem, but is complex. At least it's a right triangle.
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