Hello, camherokid!
I have a slightly different approach to motion problems.
I think it's neater ... see what you think.
At noon, Ship A is 100km west of ship B.
Ship A is sailing south at 35km/h
and ship B is sailing north at 25km/h.
How fast is the distance between the ships changing at 4 PM? Code:
* B
/ ↑
/ ↑ 25t
/ ↑
P * - - - - - / - - - * Q
↓ x / :
↓ / :
35t ↓ / : 35t
↓ / :
↓ / :
A * - - - - - - - - - * R
100
Ship A starts at point P and sails south at 35 km/hr.
. . In t hours, it has sailed 35t km south to point A.
Ship B starts at point Q and sails north at 25 km/hr.
. . In t hours, it has sailed 25t km north to point B.
Let x = AB.
In right triangle BRA, we have: .x² .= .(60t)² + 100² .= .3600t² + 10,000
Differentiate with respect to time: .2x·(dx/dt) .= .7200t
. . and we have: .dx/dt .= .3600t/x .[1]
At 4 PM (t = 4), the triangle has: AR = 100, BR = 240. .Hence: x = 260.
Substitute into [1]: .dx/dt .= .3600(4)/260 .≈ .55.38 km/hr
. . Their distance is increasing.