A boat leaves a dock at 6:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 7:00 PM How many minutes after 6:00 PM were the two boats closest together

22 minutes

Explanation:

Let t be the time after 6 PM.

D be the distance

During time t, the First boat travelling south = 20t km.

Meanwhile, the second  boat travelling east is= 15t km,

Then the second boat  started 15 km from the dock, so the  distance from the  dock is= (15 - 15t) km.

The distance D between the  two boats is is expressed as = (15 - 15t)^2 + (20t)^2 = D*2

Let differentiate with respect to time  and making  dD/dt to 0 (D minimization is important),

then  2(15 - 15t)(-15) + 2(20t)(20) = 0

Therefore t = 9/25 h or 0.36 h or 22 minutes

v 154.9= 0.9 length length s little over 154.9cm

no. dot product   =0 when vecs at right angles