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a boat leaves a dock at 2:00pm and travels due south at a speed of 20km/h. another boat has been heading due east at 15km/h and reaches the same dock at 3:pm. at what time were two boats closest together?



Answer :

The two boats will be closest at 21.6 min past 2.

At t minutes past 2PM, the southbound boat will have traveled =(  20 km /hr) x  t min

= (20 km /60 min) x  t min

= (1 /3)t   km

Similarly the eastbound boat will have traveled = (15km/hr) x t min

                =( 1/4)t km

But as it started 15 km away from the dock , so the distance from the dock is

     15 - ( 1/4)t   km

Let D be the squared distance between the boats.

Also we will minimize the time period for 0 < t < 60 min

Now ,

  D = (t/3)²  +  (15 - t/4)²  

Now differentiate and set the derivative to zero.

               D’  =  25t /72  - 15/2  = 0

 Therefore      

 t = 21.6

So for    t = 21.6

     D = 144

Or         (distance)² = 144   km²

        distance = 12 km

So  The two boats will be closest at 21.6 min past 2.

Learn more on distance-time problems here :

https://brainly.com/question/18798899  

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