Finished with school!..for now, graduation tomorrow Options

[jazzcornet]

the more i think about it, the more i need to clarify the velocity thing. the fast boat travels at an indicated 3mph (meaning 4mph true away from the dock and 2mph true towards the dock after you account for current velocity vectors). the slow boat indicates [3-(t/4)]mph. then you need to account for the current in both directions.

HTH


edit: oh what the heck, if anyone else can get this problem right with all the work shown, i will gladly give you 50% off as well. only the first three people to submit the correct response will receive this deal

This post has been edited by jazzcornet: Jun 17, 2008 - 4:14 PM

[lubu]

Ok here is my attemp.

"The current flows at a constant 1mph in the direction away from the dock" means the V(fastboat)=4mph and V(slowboat)=4-(t/4)mph
To find the distance from origin(dock) at any t we must integrate V

R(fb)=4t + c , R(sb)=4t-[(t^2)/8] + c

c is not important in this case

after one (t=1) :

R(fb)= 4(1) = 4 miles
R(sb)=4(1)-[(1^2)/8]=3.875miles

The fastboat then turns around heading back for the dock, when it turns around the new V(fb)=2mph and R(fb)=2t
the fastboat intercepts the slowboat at ~t=0.0208 (which is abt 1.25 mins after it turned)

The slowerboat total distance travelled is:
t=1.0208
R(sb)=4(1.0208)-[(1.0208^2)/8]=3.9529 miles

Now the slowboat turns around and heading back to the dock, new V(sb)=2-(t/4) and R(sb)=2t-[(t^2)/8]
We need to find the time, we've got the distance travelled:
3.9529=2t-[(t^2)/8]
Reaareange this we get the quadratic fomula then solve for roots ---> 0=-[(t^2)/8]+2t-3.9529
t=2.3099 and t=13.69
we can discard t=13.69

t=2 means 2 hours and the 0.3099 = ~18.6 mins

The slowboat arrives at the dock at 5.18pm

[jazzcornet]

your answer is elegant, but ultimately incorrect. how did you calculate at which point the boats intercect? what are the limits for your initial integration?

very good effort though, you are on the right track!

edit: i meant the limits of your second integral, but i see you didn't get to that step.....carry on

This post has been edited by jazzcornet: Jun 17, 2008 - 4:23 PM

[lubu]

your answer is elegant, but ultimately incorrect. how did you calculate at which point the boats intercect? what are the limits for your initial integration?

very good effort though, you are on the right track!

Haha i knew thats where i went wrong, I didnt bother with the limits and I kinda guessed the intercept time. Its 5am in here in Australia and Im just abt to sleep...been staying up watching euro 2008

[jazzcornet]

that was pretty close though!!! i'd love to see if you could get the right answer. i think i will post that as my riddle question for the month on my website

http://www.gsspowdercoating.com