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

[celicamike]

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.

[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