Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
swin2013
Group Title
A particle moves along a line so that at any time t> (or equal to) 0 its velocity is given by v(t) = t/1+t^2. At time t = 0, the position of the particle is s(0) = 5. Determine the position of the particle at t = 3.
 one year ago
 one year ago
swin2013 Group Title
A particle moves along a line so that at any time t> (or equal to) 0 its velocity is given by v(t) = t/1+t^2. At time t = 0, the position of the particle is s(0) = 5. Determine the position of the particle at t = 3.
 one year ago
 one year ago

This Question is Closed

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
i know i must take the antiderivative
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
yea i did u sub for 1 + t^2
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
i meant +
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
i got 1/u du
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Woops, should be a factor of 1/2 in front I think.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
and i should get u = 1 and u = 10?
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
oh yea 1/2 (1/u) du
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
No we don't want to add limits to our integral.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Evaluate the `Indefinite Integral` to get a function of position, s(t). It will include an unknown constant +C. We can use our `Initial Conditions` to solve for C.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
oh wait... i do the t(3) = t(0) + the integral
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
well i did something similar... but i put t(0) _
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
so i plug 3 for t?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
After you've determined the value of C, yes, the final step is asking for s(3). So you would plug in t=3 and solve.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
i gotsss 3.848
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
idk if that's right.... scary
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hmm I got 6.15
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
oh i subt 5 _
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.