How to find r(t) by given:v(t) r(2) ?

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AI Thread Summary
To find the position vector r(t) from the given velocity vector v(t) and the known position r(2), the integration of v(t) must be performed with appropriate limits. The equation r(t) = r(0) + ∫(v(t)) dt indicates that the initial position r(0) is necessary for calculating r(t). Without r(0), it is challenging to determine the exact position at any time t, including t=2. The discussion highlights confusion regarding how to utilize the provided r(2) and the implications of not knowing r(0). Understanding the relationship between the initial position and the integral of the velocity is crucial for solving the problem effectively.
Lenjaku
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Homework Statement



Calculate the position vector at any given moment ( r(t) )

v(t)=(6t^2+4t)i+(3t-2)j+(12t^3-t^3)k m/s
r(t=2)= (3,-0.5,-2)

Homework Equations



r(t) = r0+ integral(v(t))

The Attempt at a Solution



I tried to integrate v(t) and get r(t) but I don't know how to calculate r0.
what am I supposed to do with t=2 I was given?
if I do integral of v(t) and calculate the it by t=2 - t=0, what would be the meaning of this?
t=0 doesn;t have to give r(0)=(0,0,0)

We learned to get the vectors with known starting points so I am puzzled about finding them myself and unfortunately all the questions deal with finding them -.-
I seek to learn the meaning of the whole picture.
then I can solve the rest.
 
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Welcome to PF!

Hi Lenjaku! Welcome to PF! :smile:

In the formula r(t) = r(0) + ∫ (v(t)) dt,

the limits of integration have to be 0 and t …

r(t) = r(0) + ∫0t (v(t)) dt​

But you can substitute any value a of t …

r(t) = r(a) + ∫at (v(t)) dt :wink:
 
Unfortunately I am still trying to figure out things we only had 1 lecture about this.
If I take the integral of t=2 and reduce t=0 it woudlnt give me anything (I think).

Just to see what happens I integrated v(t) and placed t=2. the numbers are nothing near r(2).

it turned out to be:
∫v(t)=24i+(-1)j+12k
r(2)=3i-0.5j-2k

What would be the meaning of this difference?

Both is the same time,but apparently the starting point affects the result...did I calculate ∫v(2) as r(0,0,0) starting point? if so why it turned to be so if at all?
 
Hi Lenjaku! :smile:
Lenjaku said:
Just to see what happens I integrated v(t) and placed t=2. the numbers are nothing near r(2).

it turned out to be:
∫v(t)=24i+(-1)j+12k
r(2)=3i-0.5j-2k

not following you :confused:

v(t)=(6t2+4t)i+(3t-2)j+(12t3-t3)k m/s

(hmm … that last bracket looks wrong :redface:)

can you show all the steps for the way you calculated ∫ v(t) dt?
 
I integrated v(t) as:
(6t3)/3 + (4t2)/2 , (3t2)/2 - 2t,(12t4)/4 - (9t4)/4

And then placed t=2.

r(2) was given.

I simply wanted to see what it turns to be to understand what is going on there...

I still don't understand how to find r(t) at any given time when I am only given r(t=2) and no r(t=0) how do I find r(t=0) (do I need it?, if not why)?

I think I need it to put it in the equation:

r(t=0) + ∫v(t)
I have v(t) I can get its integral but what do I do with the given r(t=2)?
Why was I given this for?
 
v(t2) - v(t1) = [(6t3)/3 + (4t2)/2 , (3t2)/2 - 2t,(12t4)/4 - (9t4)/4]t1t2

you subtract the initial value from the final value
 
I am sorry I don;t get you.
The only t I was given is t=2.
And I was give the components of r(t=2) so I assume the answer should do something with it.

I don;t know what r(t=0) is.

What do I do with the r(t=2) I was given?

Doing as u suggested was my initiate approach with no any further thinking it seems logical but it leaves me with too many questions.

I am thinking of x=x0+v0t
taking v as constant. So v0=v.
 
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