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?
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
This question is so confusing. the only thing i could come up with was x is 20 (since it travels for an hour) and x' = 20km/h and y'= 15km/h.
|dw:1353476410429:dw| --> z^2 = x^2 +y^2
We implicitly derive that equation so it becomes 2zz'=2xx'+2yy'
Thats all I have until now, can anyone add to that?
The top equation represents the distance from the dock of the boat heading east at t minutes, the right one represents the distance from the dock of the boat heading south.
You can find the distance using Pythagoras' formula
This must be a minimum, and as you will notice the distance is a quadratic equation so you should be able to find the minimum :)