At noon, ship A is 150Km west of ship B.
Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h.
How fast is the distance between the ships changing at 4:00pm?
x * |
* | 25t
* - - - - * - - - - - * Q
P 35t A 150-35t
Ship A starts at point P, 150 km west of point Q.
. . In t hours, it has moved 35t km east to point A.
. . Hence: .AQ = 150 - 35t
Ship B starts at point Q.
. . In t hours, it has moved 25t km north to point B.
Let x = AB
Then: .x² .= .(25t)² + (150 - 35t)²
. . and we have: .x² .= .1850t² - 10,500t + 22,500 . 
Differentiate with respect to time: .2x(dx/dt) .= .3700t - 10,500
. . and we have: .dx/dt .= .(1850t - 5250)/x . 
When t = 4, we have: . .x² .= .1850·4² - 10,500·4 + 22,500 .= .10,100
. . Hence: .x .= .√10,100 .= .10√101
Substitute into : .dx/dt .= .(1850·4 - 5350)/(10√101) .= .21.39329959
Therefore, the ships are separating at about 21.4 km/hr.