00:01
Hi, here we have the following problem.
00:03
A train moves along a straight line.
00:04
Each location at time t is given by s of t equals 100 over t, t from 1 to 5.
00:11
T is measured in hours and s t is in kilometers.
00:16
First we're going to graph f of s of t for t between 1 and 5.
00:21
So what we are going to do an approximate graph what i'm going to try to do is just give you the correct shape of the graph.
00:33
So the shape of the graph is going to be like this.
00:37
And here we have t equals 5 and let's say here we have t equals 1.
00:45
Over here we know that s of 1 is 100, s of 5 is equal to 20.
00:57
To have 20 here and we are going to have 100 here so this is time and this is s of t okay now the first thing we okay so we did the first thing second thing says um find the average velocity of the train between t equals 1 and t equals 5 the average velocity is given by this um by this formula it is the uh displacement over time elapsed so in our case this is going to be average velocity is equal to s of 5 minus s of 1 over 5 minus 1 so this is 20 minus 100 over 5 minus 100 over 5 minus 1 so we have minus 80 over 4 is minus 20 kilometers per hour.
02:06
The minus here, the negative sign shows direction, so that means the train is moving backwards.
02:14
So velocity is something like you have to think it as a vector, that it has direction, the minus sign shows that the train is moving backwards.
02:22
The next question it says, where on the graph of eflstc can you find the average velocity? so the average velocity, as we said it is displacement over time elapsed, okay? so the displacement is here over time elapsed, and this is this vector over here.
02:49
You can also think of the average velocity as the slope of this line segment, which is downward sloping, so that's why we also have the negative sign.
03:01
Okay, so now the next thing says use calculus to find the instantaneous velocity of the train...