Ok, so I get:
1.5x^2 + 0.194x + .09 =0
I end up have to take the sq root of a negative though, so I must be doing something wrong. -
sqrt(0.194^2 -4(1.5)(.09)) = -.502
Homework Statement
A horizontal uncompressed spring of constant k= 3 N/m and negligible mass lies on a frictionless floor. The right end of the spring rests against a vertical wall. A block of mass m=0.180 kg with initial velocity v0= 1.0 m/s is incident to the spring from the left. The spring...
HELP! Diff Eqs, Torricelli's law
Hi, my crazy diff eq professor for some reason decided to assign us a project accompanied by some book problems right in the middle of exam week, due friday (day of my last exam). This is easily the most redic. thing I've ever seen in my college career. Anyhow...
Its for a course that already expects you to remember to remember this, we're just doing the quick review chapter in the beginning. I've never taken physics though, so I hardly every deal with this stuff, so I forgot exactly how to go about the problem. So are you saying then that I want to...
the problem is given a(t)=4(t+3)^2, v0=-1, x0=1 find x(t), the position function
to my knowledge x(t)= 1/2at^2+v0t+x0
I tried it and got the wrong answer, the answer the book gives is: 1/3(t+3)^4 - 37t -26
I did:
x(t)= 1/2 [4(t+3)^2)]t^2 - 1t +1
eventually getting: 4t^2 + 24t + 9...
([SIZE="2"]Moderator's note: thread moved from "Differential Equations")
The DE is y''=a*sqrt(1+(y')^2)
I have no idea how to go about integrating it, I just started taking diff eq's and haven't taken calc in over a year. Any help would be appreciated, thanks!