00:02
So here we're given a velocity function for a particle moving along a line, and that velocity function is v of t equals t squared minus 4t minus 12 over the domain t between 1 and 7.
00:19
And we want to find the displacement of the particle, as well as the distance traveled.
00:26
And so the displacement, just sort of like the net distance you are from your starting point, right? because you could be moving sort of back and forth.
00:37
And so if you move forward a little bit and backwards, you've traveled a further distance than your displacement, right? you know, if you go forward five feet and then back three feet, right? you've traveled a total of eight feet, but you're only two feet from where you started.
00:53
So there's a difference between displacement and distance traveled, and i want to look at the graph of this velocity function to really figure out what to do here.
01:03
So here's our coordinate system here.
01:09
Here's the t -axis.
01:12
And there's v.
01:15
And we care about values from 1 to 7.
01:20
I draw this in blue here.
01:22
So what this looks like, it's going to be something like this.
01:30
Just a little snippet of an upward -facing parabola.
01:33
The graph goes on like this, you know, it's a parabola.
01:38
But we only care about the piece between t -quels -1.
01:40
It equals seven.
01:42
So the graph looks something like this.
01:44
And you see that at some point, it crosses the t -axis, right? and there's like this sort of green chunk that sits below the t -axis, and then there's this red chunk here that sits above the axis.
02:07
And when you take the integral of velocity, you get this placement.
02:16
And what it does is it sort of gives you this signed area because that green region, so the integral of the function v, let me just write this down here, that green area is given by the integral of the velocity function from t equals 1 to t equals whatever that red point is.
02:41
I'm just going to fill in with a little red question mark right now.
02:43
We don't know where that intercept is at the moment.
02:48
But that integral is the area of that green thing.
02:53
But in general, this is going to be a negative number because it's below the t -axis.
03:01
So it's like a negative distance, so to speak.
03:05
So it's some sort of negative value.
03:06
Whereas this red region is the integral from the question mark, right? we don't know what that t value is up until 7.
03:19
And this thing is positive because it's above that t axis.
03:25
And so if i integrate from, right, from 1 to 7, it's going to give me a much of negative area for the green part plus the positive area.
03:36
Right.
03:37
And so sometimes people call this signed area.
03:39
Again, this gives you the displacement because it's going to add the positive bit to the negative bit or subtract away that negative amount from the positive amount.
03:50
Now, so displacement then is just simply, right, it's like the signed area calculation.
04:01
So what it is is the bit under the curve, so the integral from 1 to this question mark, v of t plus the integral from that question mark up to seven.
04:17
U .t.
04:18
Oops, i forgot my dts.
04:20
That's just the integral from one to seven of the velocity.
04:27
That's all that is.
04:30
And what's the distance traveled? well, i just don't want to subtract away any area, right? that's going to be like me, like, i'm going to lose that information of that distance that i traveled...