# [SOLVED] Position and Velocity vector??

• March 28th 2010, 05:39 AM
Lafexlos
[SOLVED] Position and Velocity vector??
Find the position and velocity vectors if the acceleration is
$A(t)= (cost)i -(tsint)k$
and the initial position and velocity vectors are $R(0)=i-2j+k$ and $V(0)=2i+3k$ respectively.

Any help is appriciated, bla bla..
• March 28th 2010, 05:42 AM
craig
Quote:

Originally Posted by Lafexlos
Find the position and velocity vectors if the acceleration is
$A(t)= (cost)i -(tsint)k$
and the initial position and velocity vectors are $R(0)=i-2j+k$ and $V(0)=2i+3k$ respectively.

Any help is appriciated, bla bla..

Show us what work you have done so far, we can help you from there. bla bla
• March 28th 2010, 06:38 AM
Lafexlos
Unfortunately, couldn't do anything on it.
It looks gonna take integral of acceleration and put zero and gonna find velocity but i can not take the integral of vector. :S

is it same as normal integral and put i , j , k? or does it exist something like integral of vector?
• March 28th 2010, 06:41 AM
craig
Quote:

Originally Posted by Lafexlos
Find the position and velocity vectors if the acceleration is
$A(t)= (cost)i -(tsint)k$
and the initial position and velocity vectors are $R(0)=i-2j+k$ and $V(0)=2i+3k$ respectively.

Any help is appriciated, bla bla..

To integrate a vector all you do is integrate the individual components.

For example, a vector such as $t^2 i + 3t j - 5k$, differentiated with respect to $t$ would be, $2t i + 3 j$.

Hope this helps.

Edit: The same rules for integration apply.
• March 28th 2010, 06:42 AM
skeeter
Quote:

Originally Posted by craig
Show us what work you have done so far, we can help you from there. bla bla

$a(t) = (\cos{t})i - (t\sin{t})j$

i component ...

you should already know the antiderivative of $\cos{t}$.

j component ...

you'll have to use integration by parts to find the antiderivative of $t\sin{t}$. give it a go.
• March 28th 2010, 06:57 AM
Lafexlos
So, $\int A(t)= (sint) i+ (c) j + (sint-tcost) k + C$
i think, now i should put zeros instead of $t$ but when put zero for $i$ component i get $0i$ but instant velocity has $i$ component. Does it mean that $C$ has $i$ component?
Bah. Really confused but understand the logic. =) Thx for helps.
• March 28th 2010, 07:09 AM
HallsofIvy
You have forgotten the "constants of integration" for each component.
• March 28th 2010, 07:12 AM
Lafexlos
Ok. I see.
So it'll be $(sint + C)i$ instead of $(sint)i + C$. Now everything is complete.