Question

integral velocity question!
v(0) = 1, v(1) = 2, v(2) = ?

Answer 1

a = a(t)m/s^2
v = v(t)m/s

v(0) = 1
v(1) = 2

\[a = k \frac{ t^{3} }{ v^{2} }\]

Find v(2)

Answer 2

I was thinking this.\[v(t) = \int\limits\limits_{}^{}a(t)dt+C\]Because k is a constant already being used.

Answer 3

I am a little stuck when looking at the acceleration formula we have to integrate.\[v(t) = k\int\limits\limits_{}^{}\frac{ t^3 }{ v^2 } dt+C\]The v^2 is tripping me up... hmm

Answer 4

How would I use this equation though? Do I just plug the values in?

Answer 5

I still can't figure out what the constant k is with what I'm given

Answer 6

After we integrate we have two equations we can setup because of the initial conditions that were given. With these 2 equations we can solve for both of our unknowns.

We just have to figure out how to integrate this first. What class is this?

Answer 7

its a calculus ii course

Answer 8

So have you learned integration by parts yet?

Answer 9

yup

Answer 10

I know how to do it

Answer 11

I think that is what is needed here

Answer 12

write v as v(t). Now both unknowns are in terms of t

Answer 13

you mean replace v in the equation with v(t)?

Answer 14

\[v(t) = k\int\limits\limits_{}^{}t^3v^{-2}(t)dt\]

Answer 15

Ya both need to be in terms of what you are integrating

Answer 16

Then apply integration by parts

Answer 17

Ya. I'd let t^3 be u and v^-2(t) dt be dv

Answer 18

ok, I'll work on it and see what i get, give me a few minutes to calculate

Answer 19

It didn't get me anywhere. This might be one of those annoying ones where you have to do it twice.

Answer 20

k

Answer 21

After one IBP
\[t ^{3}-\frac{ 1 }{ v(t) }dt - \int\limits_{}^{}-\frac{ 1 }{ v(t) }dt 3t ^{2}\]

Answer 22

\[t^3(\frac{ -1 }{ v })+\int\limits_{}^{}v^{-1}(t)3t^2dt\]

Answer 23

\[uv-\int\limits_{}^{}v \space du\]

Answer 24

for second IBP, which should I choose for u and dv

Answer 25

But this gets us nowhere. Even after you do it twice. I'm going to rethink this problem.

I used the same things again for taking it twice. if you switch it will just go back to what it was.

Answer 26

I'm calling in a consultant @zepdrix

Answer 27

Hmmm it feels like you guys were on the right track at the start... This info should help us find k in some way I think.. Hmmm
\[\huge \int\limits_{t=0}^1 a(t)dt=v(t) \; |_0^1\]
\[\huge =v(1)-v(0)=2-1\]

Answer 28

Oh maybe I'm not doing that right hehe XD thinking.

Answer 29

I see what you did. I thought of that but we still need an equation still to find v(2).

Tough question @LockeMcDonnell

Answer 30

wait what do you mea nby regular integral?

Answer 31

like solvable?

Answer 32

Nvm. I still can't get rid of the v, hmm, I hate this.

Answer 33

I have to go. Sorry to leave this unsolved. :(

Answer 34

no problem chme, thanks for your help though