00:01
All right, so here we're going to look at a object moving with a motion graph or position graph of f and t.
00:12
And the position with time on the straight line is t cubed minus 8 t squared plus 24 t, where t is in seconds and s is in feet.
00:24
And we're only interested for t greater than or equal to zero.
00:27
And so what we're going to do is based on this motion equation, we are going to solve, answer a lot of questions.
00:34
So let's go to it.
00:36
Okay, first of all, we want to find v of t.
00:38
And v .a t is the derivative of our position graph.
00:42
So we're going to go ahead and use power rule.
00:44
We'll get 3t squared minus 16t plus 24 by power rule.
00:50
And our units will be feet per second.
00:53
So in feet per second.
00:55
All right.
00:56
So in part b, we want the velocity at one second.
01:00
So that would be just what happens when you plug in one for t.
01:04
So we'll go ahead and plug in one.
01:07
And that will give us three minus 16 plus 24 or 11 feet per second.
01:14
So that is, i'll just box our answer so far.
01:19
All right, let's take a look at part c.
01:21
Part c is we want to find out when our object comes to rest.
01:27
So we need v of t, which is 3 t squared minus 16 t plus 24 to equal zero.
01:36
And then the goal is to solve for t.
01:39
So not necessarily easy to factor when you have that three in front.
01:44
So let's go ahead and use quadratic formula.
01:47
So quadratic formula says then t will be equal to minus b.
01:52
You remember we've got a, b, and c.
01:56
So t is minus b, so 16 plus or minus square root of b squared, so minus 16 quantity squared, minus four times a times c.
02:11
And that's all over 2a, so over 6.
02:14
The trouble is, is that when we plug in to our calculator, we end up, i just kind of, leave room here we get 16 plus or minus oh no a minus 32 square reading a negative gives us an imaginary answer therefore there are no zeros that are real therefore vf t never uh goes to zero therefore we can say um the object never uh is at rest okay so uh vt never goes to zero okay so that's good to know all right, let's keep going up top.
02:57
Next thing we need to know is when is the particle moving in the positive direction.
03:03
Well, positive direction corresponds to velocity being positive.
03:11
Well, i'll just say being positive, not just positive.
03:15
All right, since our velocity, we know the object is never at rest.
03:20
We can just plug in any time.
03:21
And if we get a positive value, it's always that way.
03:25
So let's plug in t is zero and it's positive.
03:29
So since v of zero is greater than zero and we never stop, that means we're always going in the same direction.
03:40
So therefore, we are always, the object is always going in the positive direction.
03:53
Okay, good to know.
03:55
All right, let's keep going.
03:59
We solve some more to go.
04:00
All right.
04:01
So part e, we need to find a total distance.
04:05
And since we're always going in the same direction, the total distance is really just the change in our position.
04:12
So position at, and we're doing it over the first six seconds.
04:18
So the difference between where we're at six seconds and where we're at zero seconds.
04:22
So at six seconds, we plug in six -story equation.
04:27
So we're going to get six cubed minus eight times six squared plus 24 times six.
04:34
And then all the terms have t in it.
04:37
So s of zero is zero.
04:39
And when we plug that in our calculator, we end up with 72 feet.
04:44
So you traveled 72 feet in those first six seconds.
04:49
So this is for six seconds.
04:51
First six seconds.
04:57
Okay, excellent.
04:57
All right, we can keep going.
04:59
We have lots of parts...