Asked by mariel
A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 15 km/h. Another boat has been heading due east at 20 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?
Answers
Answered by
Steve
as usual, draw a diagram. The distance z between the boats at time t hrs after 2:00 is
z^2 = (15t)^2 + (20-20t)^2
2z dz/dt = 2(15t)(15) + 2(20-20t)(-20)
z dz/dt = 1250t - 800
clearly dz/dt=0 when t=8/1250 hours
Now just change that to a time of day.
z^2 = (15t)^2 + (20-20t)^2
2z dz/dt = 2(15t)(15) + 2(20-20t)(-20)
z dz/dt = 1250t - 800
clearly dz/dt=0 when t=8/1250 hours
Now just change that to a time of day.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.