The velocity of a particle is given by v(t)

= (t^2) i+ (t^3 +1) j

and the particle is at

the point (2,1) when t=0.

a) Where is the particle when t=2?

This was on a sample final and gave no answers...

thanks!

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- May 5th 2009, 08:15 PMiamtrojan3A relatively easy Vector problem from calc 3?
The velocity of a particle is given by v(t)

= (t^2) i+ (t^3 +1) j

and the particle is at

the point (2,1) when t=0.

a) Where is the particle when t=2?

This was on a sample final and gave no answers...

thanks! - May 5th 2009, 08:34 PMTheEmptySet
Remember that velocity is the derivative of position

$\displaystyle \frac{dr}{dt} = (t^2) \vec i + (t^3+1)\vec j$

So we take the integral

$\displaystyle \int dr = \int (t^2)dt \vec i + \int (t^3+1)dt \vec j$

$\displaystyle \vec r(t) = (\frac{1}{3}t^3+c) \vec i + (\frac{1}{4}t^4+t+d)\vec j$

$\displaystyle \vec r(0) = 2\vec i+\vec j=(\frac{1}{3}(0)^3+c) \vec i + (\frac{1}{4}(0)^4+0+d)\vec j$

so we get c=2 and d=1

$\displaystyle \vec r(t) = (\frac{1}{3}t^3+2) \vec i + (\frac{1}{4}t^4+t+1)\vec j$

$\displaystyle \vec r(2)=...$ - May 5th 2009, 09:00 PMiamtrojan3
So this was my reasoning, not quite the same but it comes out with the same answer as you

To get the position at some later time we must integrate the velocity over time since distance is velocity times time. So at some later time the position will be given by:

P = A + Integral{V*dt}| from 0 to 2

So at time t = 2 we get:

Integral {(t^2, t^3 + 1)}dt = [(t^3/3), (t^4/4 + t)]

P = (2,1) + (2^3/3, 2^4/4 + 2)

P = (2,1) + (8/3, 6) = (14/3 , 7)

So the particle is at (14/3 , 7) when t=2

can you check if thats correct?

sorry i don't really no the coding for this...

And, thanks alot, really appreciate your help. - May 5th 2009, 09:04 PMTheEmptySet
Yes you answer is correct and here is a link for the LaTex code

Help:Displaying a formula - Wikipedia, the free encyclopedia

Is is really nice and not too hard to learn.

Just hit the sigma button to open the math tags.