00:01
All right, so what we have here is a more challenging problem just because there's so many time intervals that are occurring.
00:11
But if we draw a really, really careful diagram, then that will help us not lose track of what's going on.
00:18
So i think it would be most useful to just use a really, really long number line for this problem.
00:26
Because it's one -dimensional.
00:28
We're only horizontal, we're only the x -axis, so we don't need to draw a whole graph.
00:31
We don't need to worry about a vertical or y direction.
00:33
We're just going to draw a number of y to represent our x -axis.
00:38
So we can draw a big dot right here to indicate that that's our starting point.
00:43
That's where the puck is, and our puck starts at a location of x -not is equal to zero, as well as time is equal to zero.
00:54
So we'll draw that in blue.
00:55
So we'll say that t not is equal to zero and we're going to travel along to three different time intervals so um the information that we've realized so far is that it starts off at x equal zero and t equals zero i'm just reading ahead we're going to cross by three other times so those are two seconds five seconds and seven seconds um i propose naming those by t1, t2, and t3 respectively.
01:32
So t1 is going to be equal to two seconds.
01:35
And just to save space, i'll not worry about the significant figures on my diagram, but t2 is going to be equal to five seconds.
01:44
And then likewise, t3 is going to be equal to seven seconds.
01:53
And then we don't know.
01:55
I kind of drew them evenly spaced here, not to imply that they're all going to be evenly spaced, but these are technically unknown distances as of yet that we have to solve for throughout the problem.
02:08
So we'll just follow the subscript convention, and column x1, x2, and x3.
02:19
So this is our schematic of what's going on in the problem, and then the reason why the puck is even moving in the first place is because they're may or may not be a force acting on each of these time intervals that i'm drawing in green right here.
02:35
So what that means is that we have a non -zero force.
02:40
So that is this 0 .25 newton's.
02:45
So in the first time interval, the force is being applied.
02:55
And then again, with that we're about sitting the figures, 0 .25 newtons.
02:59
But in the second time interval, the force is no longer being applied, so that gives you a force of zero.
03:07
But then in the third interval, the forces again turned on because the player is pushing on again.
03:14
So we again have a force of 0 .25 newton's going on here.
03:20
So those are the complexities that are going on.
03:24
We can already see if the force is going to be equal to zero, then newton's second.
03:30
Law which is f is equal to m a is going to tell us that the acceleration is equal to zero um but the acceleration is going to be non -zero in the first and the third interval but anyway i'm getting a little ahead of myself so oops um it's a circle that in a different color in part a we will be solving for this position as well as its respective velocity and and then part b, we're going to be solving for x3, the final position and velocity.
04:08
So this is our schematic diagram of what is going to go down in this problem.
04:15
And so let's just go ahead and get started with mathematically with part a.
04:20
Now that we have our very carefully crafted diagram, so we're actually going to be using kinematics equations.
04:27
In case you can tell, in order to solve for the positions and velocities.
04:34
So x1 is going to be equal to the initial position x not added to the initial velocity v knots.
04:50
And sorry, these are supposed to be parentheses.
04:56
The time interval from t1 to t0, it's gonna be super explicit here, plus one half of this acceleration from t1 through t0 squared is important on the acceleration term.
05:20
So you can already see here one other piece of information that was given to us that i'm about to sketch out is that we also know that the puck begins at so in other words, its initial velocity is equal to zero.
05:31
That really helps us out on this first equation right here, because velocity is equal to zero, so that second term goes to zero.
05:41
And in addition, it starts to the origin, so its initial position is equal to zero.
05:46
So that will help us out there, so we only need to plug in numbers for the third term.
05:53
So we have a one half at front, an acceleration that is going to be equal.
05:58
To 0 .25, 0 divided by 0 .16, um times this time interval to 0 .00 minus 0 squared, which to three significant figures will give us 3 .13 meters.
06:33
Great.
06:36
So that's the first half of part a.
06:41
I think we have enough space down here to squeeze in the velocity.
06:47
So we can use a simpler kinematics equation, that being v1 is equal to v0.
06:54
Simply added to the product of our acceleration multiplied by the time interval that we're working over.
07:03
Resets from t0 to t1, no squared, n 1 half.
07:07
Just like before, we know that the initial velocity is zero, so that first term goes away.
07:13
So we are just left with that same acceleration, which is force divided by mass.
07:33
There's our acceleration, and then our time interval, square or anything like that.
07:43
And then at the end of the day, sorry, it's starting to look at checkmark, v1 is, it gives you the same answer because of the coincidence that the time interval is two seconds.
07:54
It was squared before, but it canceled out with the one -half of the friends.
07:57
This is also numerically 3 .13, but it's a velocity, so it has different units.
08:03
It comes with meters, units of meters per second and not anything else.
08:08
Okay...