Well, its over. I just finished my last final exam this morning, and I dont have anything else to do other than walk up onto a stage and grab my degree. It's been a long journey, but tomorrow I will be graduating from Cal State San Bernardino with a B.S. in Chemistry (with honors) . I wish I was done with school forever, but it's going to be another 4-6 years because I've been accepted to graduate school. I normally don't like to blow my own horn, but you'll have to excuse me because I'm just feeling really stoked right now.

It would be nice to get a full time job and put the kind of money into the celica that I've been wanting to for the last three years as a broke college student. In the long run though, I'd like to futher my education. My celica's been a good college car, and I'm hoping it will last me through grad school, because I'll be going from being a broke college student to a broke grad student. At least I don't have any debt and I own my car outright.

Now that school's over I'll need to start looking for a place to live in LA, but first, I'll be taking a week long camping trip with some friends that we've been planning for a while now. I've been looking forward to this for some time, and I'm just feeling really good right now. Thought I'd share.

Congrats

I can't wait to be done with school.

in honor of your education i'd like to offer you a 50% off discount for powdercoating services for answering a simple math question. are you up for the challenge?

I finished my last final yesterday and went straight to the clubs glad it's over for now. I would be graduating now but I chose ME which takes 5 years for most people . congrats on graduating

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So your saying you have access to a CNC machine and machinist? . Bunch of my buddies are engineering students here at the university of delaware, tough program but excellent degree to have... always a need for engineers.always

This post has been edited by laff09: Jun 14, 2008 - 2:00 PM

post your engineering degrees:

jazzcornet - B.S. Nuclear Engineering m. Theoretical Mathematics / MIT

jazzcornet.. you sir... must be amazingly smart.

im just finished my first year of compsci engineering and i must say that i hate it and i want to shoot myself and im probably going to transfer to another engineering... haha

congrats to celicamike tho i cant wait for three more years and be able to graduate... haha

Well, the ceremony yesterday was great, and then afterwards we went out to eat and I had an awesome steak dinner; it was a good day.

Father's day is today so I'm going to let my dad know how much I appreciate him.

Thanks to everyone on here for your replies, and congratulations to all the other graduates!



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Thanks for the offer, but I'm not interested in powdercoating. Just out of curiosity though, I would like to see the math problem (now that I'm out of school, I don't have anything to do) lol.

Two canoes leave a pier at 2pm. The fast boat travels at a constant 3mph. The slow boat travels at (3-t/4)mph. After an hour, the fast boat turns around and heads back to the dock. When the fast boat intercepts the slow boat, the slow boat turns around. The fast boat returns at 5pm. The current flows at a constant 1mph in the direction away from the dock. At what time does the slow boat return? Please show all your work

wow i hated algebra~ hah

congrats on the graduation tho~!

I came up with an answer but it doesnt seems right...anyways the slower boat travels at (3-t)/4 or 3-(t/4)?

i have the formula/ diagram thingy in my head but im too lazy to solve it.... lol

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Just to clarify, the speed of the boats are water speed, and not land speed right. (or velocity, whatever lol)

Congratulations, the best years of your life are now over

Enjoy the the rest of your life. It's all downhill from here.

sorry for the late replies guys, i've been busy powdercoating.

for clarification:

>>V>>>>slow boat>> = (3- (t/4))mph

and the velocities are relative velocities, which implies that you need to factor in the current as well.

good luck!!

here's the only hint i'll give you:

find the limits of integration first and save yourself the headache

This post has been edited by jazzcornet: Jun 17, 2008 - 1:54 PM

<------ BFA in Graphic Design=let the Mac do the math for me because my math intellect has been in a coma since 1991. But if you show your work, I'd be happy to design it in an aesthetically pleasing fashion!

Oh how I envy you brainiacs, what with your intellectual capacities and what-not. Jazz, let's make really smart semi-jewish babies together!

But I don't want smart jewish babies. I want a good ol fashioned child prodigy so I don't have to work ever again!!!

>

hrms...weird

3t and (3-(t/4)) if you sub t=0

ull get o and 3 then initially the slower boat is 3mph ahead?

edit: wrong expression

This post has been edited by lubu: Jun 17, 2008 - 2:29 PM

Where did you get that idea??

nvm...I think I need to eat..

This post has been edited by lubu: Jun 17, 2008 - 2:49 PM

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

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

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

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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

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