00:01
All right, we've got this position function, given time, starting at zero, and in part a, we want the velocity function.
00:10
The velocity function is the derivative of the position function, so let's jump into that.
00:16
Let's call this part a and this part b, so the derivative would be a prime b plus a, b prime.
00:24
So let's calculate that.
00:26
We're going to have 3 times t minus 2 squared times t minus 6 plus t minus 2 cubed times the derivative of b prime decreasing this exponent to 0 would make that part just 1.
00:45
So we can do some algebra on this by taking a t minus 2 squared out and that leaves me on the inside with 3 2.
00:56
Minus 18 plus t minus 2, which is going to give me t minus 2 squared times 4t minus 20, and that's going to give me 4 times t minus 2 squared times t minus 5 as my derivative here.
01:17
So here's my velocity function right there.
01:19
Part b wants to know when does the particle stop? when does particle stop? well, the particle stops when the velocity is zero.
01:33
So if we set this equal to zero, we get t equals two with a multiplicity of two, and that's going to be important here in a minute, and then t equals five with a multiplicity of one.
01:51
So the follow -up to part b, part c, wants to know, does the particle stop, or does the particle change direction at all of its stops? and the answer here is no.
02:07
The particle only changes directly when the sign of v of t changes.
02:26
And if we think about the graph of v of t, which is a polynomial function, it's got a t intercept here, and it's got a t intercept here at 5, but the multiplicity of 2 means that this graph has a bouncing behavior at this intercept.
02:51
So the leading coefficient would be positive 4, so it's going to be up over here.
02:55
The degree overall is three, so it'll be down over here.
02:58
So, you know, this graph looks something like this without using a graph in utility, just really quick.
03:06
So we can see that there is no sign change here, which means that our function, whatever it is, is, you know, decreasing and still decreasing and then starts increasing right here.
03:24
And i don't know how far up or down it is, but, you know, that's got to be what it is right there.
03:30
So, moving on to part d.
03:36
Let's finish this up over here.
03:41
Changes direction at t equals five.
03:53
In part d, we want the particles displacement over the interval from 1 to 6.
04:05
So here we're just going to look at where it starts.
04:08
1 where it ends in 6 and then we're going to subtract that.
04:14
So we have s of 1, which would be 1 minus 2 cubed times 1 minus 6.
04:23
So 1 minus 2 is negative 1 cubed, which is negative 1, and 1 minus 6 is negative 1, and 1 minus 6 is negative 5.
04:30
So our position here is 5 at time 1.
04:35
And then let's plug in a 6.
04:37
We're going to have, let's see, 6 minus 2 cube times 6 minus 6 here.
04:52
So 6 minus 2, it doesn't matter what this is.
04:54
This is 0 times something.
04:55
So this comes out to be 0.
04:58
So our displacement is going to be 5, 5 units...