Question

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.

Answer 1

@zepdrix

Answer 2

i know i must take the antiderivative

Answer 3

\[\large s(t)=\int\limits v(t) dt\]Having trouble with the integral? It looks like another nice easy `U substitution`.

Answer 4

yea i did u sub for 1 + t^2

Answer 5

i meant +

Answer 6

i got 1/u du

Answer 7

\[\large \int\limits \frac{1}{u}du\]Yep, sounds good.

Answer 8

Woops, should be a factor of 1/2 in front I think.

Answer 9

and i should get u = 1 and u = 10?

Answer 10

oh yea 1/2 (1/u) du

Answer 11

No we don't want to add limits to our integral.

Answer 12

no?? :O

Answer 13

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.

Answer 14

oh wait... i do the t(3) = t(0) + the integral

Answer 15

So the integral gave you something like...\[\large s(t)=\frac{1}{2}\ln(1+t^2)+C\]Yes? :O

Answer 16

yes!

Answer 17

Then just plug in the `Initial Condition` to solve for C!\[\large s(0)=5 \qquad \qquad \rightarrow \qquad \qquad 5=\frac{1}{2}\ln(1+0^2)+C\]

Answer 18

OHHHHH

Answer 19

well i did something similar... but i put t(0) -_-

Answer 20

heh :D

Answer 21

so i plug 3 for t?

Answer 22

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.

Answer 23

i gotsss 3.848

Answer 24

idk if that's right.... scary

Answer 25

Hmm I got 6.15

Answer 26

:((((((

Answer 27

\[\large C=5\]\[\large s(3)=\frac{1}{2}\ln\left(1+3^2\right)+5\]

Answer 28

oh i subt 5 -_-