00:01
In this problem, we have a group of people, and i'll just draw it out to illustrate things.
00:14
And so there's space at a distance l apart.
00:20
People all have a depth of d.
00:23
So this is not to scale.
00:30
And we know the values of these things.
00:32
So l is 1 .75 meters, and d is 0 .25 meters.
00:47
Okay, and we're also told that people are all moving up.
00:50
A speed equal to 3 .5 meters per second.
01:00
And so what's happening is this sort of line of people is approaching this door.
01:06
And once they run into the door, they slam into it, they get stuck.
01:12
And so if we think about a, what's going on here? the people move towards the door, they get stuck and we get this layer of people that build up and increases in thickness at the door.
01:26
And so a is asking us to calculate the rate of increase of the layer of the people at the door.
01:33
Let's just think about what's meant by a rate here.
01:39
So we're increasing by some unit size for a given amount of time.
01:48
Right, so we know that time is going to be in the denominator and some sort of size parameter.
01:57
It's going to have to go in the numerator.
01:59
And so let's just think about this for a second.
02:03
So the units that we're adding to the layer are going to be the thickness of the people, d.
02:11
So if we plug in d for the numerator, we're saying that the rate of increase of thickness of this layer is equal to units of d per time.
02:22
And so that's how we're going to want to write this out.
02:27
And so again, we're given the value of d, we know that the value of the value.
02:31
0 .25 meters, but we're not given the value of t.
02:35
So let's write out the definition of speed, but i'm going to be very careful to write it out in terms of the context of this problem.
02:45
So when we have a speed, we know that it's a distance per time.
02:52
So we're going to throw a time in the denominator.
02:54
And i just want to think about, we just want to think about the relevant distance to put in the numerator here...