00:01
Okay, so we have that the velocity of this function is 3 times t minus 1.
00:12
So the key to doing this problem is to realize a couple things.
00:17
First thing is that d x, that your position function, derivative of your position function, is velocity, and the derivative of the velocity function is the acceleration.
00:32
Okay, so let's compute these functions.
00:35
First, acceleration is easy.
00:37
Note that the acceleration is the derivative of the velocity function.
00:42
So the derivative of 3 times t minus 1 is just 3.
00:48
Okay, and then what's the, what is the position function? well, we don't need that right now.
01:00
Okay, but we can use it to find total distance.
01:04
So first thing is for a, the minimum acceleration, the acceleration is always 3.
01:10
So the minimum a of t is just equal to 3.
01:15
So there's no real minimum.
01:17
It's always 3.
01:19
So that's the first answer.
01:22
The second thing is the total distance.
01:26
So how do you calculate the total distance? the total distance is from 0 to 5 of the absolute value of v of t dt.
01:42
Okay, so now we need to figure out what is the absolute value of v of t.
01:46
So note that v of t is bigger than 0 for t bigger than 1, and then v of t is less than 0 for t less than 1.
02:05
So you can split this total distance as integral from 0 to 1, v of t dt plus the integral from 1 to 5, v of t dt.
02:26
Okay, now v of t is bigger than 0 for t bigger than or equal to 1.
02:30
So that one just stays.
02:34
So this was going to be a total from 1 to 5 of v of t dt.
02:40
But since it's less than 0 for t less than 1, then you get a minus v of t in the other one.
02:48
So you get 0 to 1 and then you get minus, because that's what absolute values does.
02:52
It always makes the, always makes the function positive.
02:56
So v is negative...