00:01
All right, so we are given a force acting in the horizontal direction, and it has the form beta minus alpha t.
00:09
And we're told that beta is 4 newtons, and alpha is 6 newtons per second.
00:17
And this is acting on a particle with an object with a mass of 2 kilograms.
00:22
And so we want to know, after this force is subjected on this particle, there's no friction, by the way.
00:29
What is the largest value of x reached by the block? so to do this, we need to calculate what the position is.
00:36
And to do that, we actually need to calculate with the velocity.
00:38
So the velocity as a function of time is going to be the integral from zero to t of the acceleration as a function of time.
00:47
And the acceleration as a function of time is just going to be the force as a function of time divided by the mass.
00:56
And so if we substitute in all of our equations, will have something like the integral from 0 to t of beta minus alpha t prime over m d t prime and so this comes out to be like one half or sorry beta over m times t minus one half alpha t squared over m so that's our velocity as a function of time assuming we have no initial velocity, which we're told in the problem.
01:29
And so then our position as a function of time is going to be then just the integral of the velocity as a function of time over some time intervals.
01:39
So we just need to integrate this again.
01:41
And also we're told that the initial position of this we can take to be zero.
01:45
So anyway, this is going to be one -half beta over m t squared minus one -sixth alpha over m t cubed.
01:54
So that's our position as a function of time.
01:56
And now we need to find the maximum of this.
01:59
Now, to find the maximum, we technically need to go all the way here because we know that the maximum occurs when the derivative is zero.
02:07
And what's the derivative of the position? it's just velocity.
02:10
So when is the velocity zero? well, that occurs when we have beta over m times t minus one -half alpha t -squared over m is equal to zero.
02:22
And one of these ts can cancel automatically because that's t -equals.
02:27
It's not really an interesting solution.
02:29
And so the solution to this is that t equals 2 beta over alpha.
02:37
And if we plug in numbers, or let's not plug in numbers for that just yet, because we want to find what is the position when we substitute in this value.
02:45
So what we'll have is like 1⁄2 beta over m times t squared.
02:50
So that's 4 beta squared over alpha squared minus 1 6th alpha.
02:57
Over m times t cubed and so this is going to be eight beta cubed over alpha cubed and really this alpha divides with this one so it'll just give us an alpha squared so now we you know have everything in like terms and we can combine and everything and we should get something like two -thirds beta cubed over m alpha square that's just algebraically and when we combine those, you know, this first term, we have a four on top, we have a two on bottom.
03:34
So we can just write that as a two out front.
03:36
This next term, we have an eight on top, a six on bottom, so we can write that as four thirds.
03:40
So we have like six thirds minus four thirds times this.
03:44
So it's two thirds, this.
03:44
And now if we plug in our numbers, we're given that beta is, i believe, what did i say? well, it's four newtons.
03:51
So four cubed is 64 newtons cubed, divided by the mass, which is two.
04:00
Kilograms times alpha squared and alpha was six newtons per second so this is 36 newton squared per second squared and so this maximum position this obtains if we plug all of this in should come out to like 16 over 27 meters or 0 .593 that's the maximum distances travels so that's the first part the next part of this question asked or how long does it take the block to reach this point so we already kind of have that it's two beta over alpha so the time it takes then is going to be two times four newtons over six newtons per second so just two seconds or sorry not two seconds i'm just reading my own handwriting this is four thirds of a second right eight over six is four thirds so one point one three repeating...