00:02
All right, so it looks like you have a calculus problem here where you're given the velocity function, and we need to come up with what these missing coefficients are, a, b, and c.
00:15
So we need to utilize three different points.
00:21
So point p, time is equal to zero at that point.
00:27
And then at point q, the time, i assume that's a five, but it translated from you to us as a dollar side.
00:40
So i'm going to solve with five because i assume that that's what that means.
00:46
I just misread it when it translated from your picture.
00:50
So i'm going to make that t equals five.
00:54
And the acceleration at that moment is 1 .12 meters per second square.
00:59
And then the third piece of information that gave us was that point bar, the time is 18 seconds, and the velocity at that time is 2 .96 meters per second.
01:14
All right, so how are we going to figure out what this function is? it kind of walks us through a couple of things.
01:21
So part a is basically asking us what's c in our function? well, they gave us a hint to what that has to be.
01:33
When the time is zero, if we plug that into this equation here, we get eight meters per second is what they say is going to happen.
01:44
So if i plug that into the equation, i get eight must equal a times zero squared plus b times zero plus c.
01:59
So c must equal eight.
02:04
That's first symbol or first variable.
02:10
So let's put that up here.
02:12
So this one's gonna be eight.
02:19
So one down two to go.
02:23
So part b, we have to solve for a and b using these other ideas that at point q, when t is five, a, the acceleration, is equal to 0 .12 meters per second squared.
02:46
And then at point r, it's velocities 2 .96 meters per six squared, and that happens at the 18 second mark.
02:58
So let's think of what we could do here.
03:02
If we take, here's an idea that we can use, is that the derivative of the function is acceleration at a given time.
03:21
So if we take the derivative of this, we'll have our acceleration, and then we can plug in the phi.
03:29
Let's see if that'll help us.
03:31
So we've got a derivative of v at t is 2a times t.
03:47
Yeah, it's, and this a doesn't mean the same.
03:52
I'm not, this a is not an accelerative.
03:56
It's the coefficient a.
04:00
I should use, i'm going to change these to capital letters.
04:04
I just don't want to confuse you.
04:06
These are not the same a's.
04:09
It's changing the capitals.
04:12
Just so we keep, we've got too many letters here.
04:16
B, c.
04:18
So this is going to be the a that the lowercase a is representing acceleration.
04:28
And then the capital a is this.
04:30
This coefficient on this function here.
04:35
So now we're, now we should be good.
04:38
So 2a, t plus b.
04:47
That's one of our equations.
04:49
So we can say this, if we plug in the, plug in the information, the acceleration is gonna be 0 .12 when two times a times five plus b.
05:08
When that equation comes out to be 0 .12.
05:12
So let me rewrite that into simpler terms.
05:15
0 .12 is equal to 10 a plus b.
05:19
That's one of our equations.
05:20
You have two variables on now.
05:23
Where that comes from, i did the derivative of the velocity function, which i knew.
05:29
And the derivative of that is 2a times t plus b.
05:36
And so then i plugged in what they gave us in the problem.
05:39
Problem.
05:41
Five is the time.
05:43
So t has been replaced with five and a, the acceleration at that time of five is point one two.
05:51
So there's our first equation right there.
05:54
Now we need to come up with another equation.
05:56
That's where this point r is going to come into play.
05:59
So at point r, we have this.
06:03
So at the, when the, when it's 18 seconds, the velocity is going to be two point.
06:11
So we can use the velocity equation that we have up here.
06:16
A times 18 squared plus b times 18 plus we found c.
06:30
That's a constant.
06:32
That's eight.
06:34
Right? so there's our other equation right there.
06:37
Remember when we have two variables, we have to have two equations.
06:40
So let's just go back to the easier one and and solve for b .1 .1 .2 minus 10a is equal to b.
06:55
So there's what we're going to sub in for b into our second equation.
07:03
This is our first equation here.
07:05
This is our second equation.
07:08
So now let's look here.
07:10
2 .96 equals, let's see what 18 squared is, 324a plus, plus 18 times 0 .12 minus 10a.
07:37
So i substituted in for b with what we knew b was equal to in terms of a plus eight.
07:46
So now we can solve for a.
07:48
We're only missing a.
07:50
Sorry that i'm writing kind of diagram here.
07:53
It's just the, sometimes that's what happens when i write for too long here...