00:01
All right, so in this problem, we're told that a particle can be described, its position and its velocity can be described as functions of time.
00:10
And we're given a whole bunch of velocity functions for this particle and then the initial conditions for the position at time zero.
00:19
And we're asked to find what the full position function is.
00:26
So we need to recognize that velocity is the time derivative of position and therefore we can integrate with respect to time the velocity function and then that should give us the position function and since we're integrating we'll get a constant of integration out but we can use the initial condition to find out what that constant integration is okay so let's start with the first one the first velocity function is just v equals zero so if we integrate that what's the integral of zero with respect to t, that is just 0 times t plus some constant.
01:08
So this is a notation they use f of t for position.
01:19
Okay.
01:19
And i cross that out because this is multiplied by zero.
01:23
So f at times zero equals 1.
01:29
So c must equal 1.
01:35
Clearly this equation does not depend on time because time was canceled out.
01:40
So for the first one, f of t just equals 1.
01:46
Okay, so that one was easy.
01:48
But the next ones get a little bit more tricky.
01:51
So our next function, v of t equals 2.
01:57
So if we take the time derivative of, or the time integral of this, we'll get f of t equals 2 times.
02:07
T plus c some constant of integration um but when t is zero um we know that this must equal five okay so um that that is zero and c we can see is equal to five so f of t equals two t plus five for that one okay, on to the next one.
02:44
We have v of t equals for t.
02:51
So let's take the integral of that.
02:54
F of t equals for t squared, divided by two or two t squared.
03:04
Oops, i forgot, plus the constant integration plus c.
03:12
All right, but f of zero.
03:18
We're told equals 1 here.
03:25
So 2 times 0 squared plus c.
03:32
That's 0.
03:34
So c equals 1.
03:36
Okay, so ft equals 2t squared plus 1.
03:46
All right.
03:47
On to the next one.
03:50
We get v of t equals 5 plus 6t.
03:59
So let's integrate that f of t must equal 5t plus 6 t squared divided by 2 all right and f of 0 for this one is 2 and that equal oh yep and this has a constant integration again um so that's 5 times 0 plus 6 times 0 squared over 2 um so those two terms are zero.
04:37
So this is just c.
04:41
So we can, i'll just go back here.
04:48
That's the answer for that one...