Answer:
Both the boats will closet together at 2:21:36 pm.
Step-by-step explanation:
Given that - At 2 pm boat 1 leaves dock and heads south and boat 2 heads east towards the dock. Assume the dock is at origin (0,0).
Speed of boat 1 is 20 km/h so the position of boat 1 at any time (0,-20t),
Formula : d=v*t
at 2 pm boat 2 was 15 km due west of the dock because it took the boat 1 hour to reach there at 15 km/h, so the position of boat 2 at that time was (-15,0)
the position of boat 2 is changing towards east, so the position of boat 2 at any time (-15+15t,0)
Formula : D=[tex]\sqrt{(x2-x1)^2+(y2-y1)^2}[/tex]
⇒ [tex]D = \sqrt{20^2t^2+15(t-1)^2}[/tex]
Now let [tex]F(t) = D^2(t)[/tex]
∵ [tex]F'(t) = 800t + 450(t-1) = 1250t -450\\F'(t) =0[/tex]
⇒ t= 450/1250
⇒ t= .36 hours
⇒ = 21 min 36 sec
Since F"(t)=0,
∴ This time gives us a minimum.
Thus, The two boats will closet together at 2:21:36 pm.
