# Answer on Calculus Question for hsd

Question #18719

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 25 km/h. Another boat has been heading due east at 10 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?

Expert's answer

Assuming that t - time in hours distance between these boatsis expressed by s(t)=sqrt(100(1-t)^2+625t^2)

We need to find the minimum of this function. We find the zeros of it's

derivative:s'(t)=1250t-200(1-t)/2*sqrt(100(1-t)^2+625t^2)=0 when1250t-200+200t=0 t=100/725=4/29 which equals 8.3 minutes.

We need to find the minimum of this function. We find the zeros of it's

derivative:s'(t)=1250t-200(1-t)/2*sqrt(100(1-t)^2+625t^2)=0 when1250t-200+200t=0 t=100/725=4/29 which equals 8.3 minutes.

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