00:01
In this prong, we're given the position as a function of time of particle a and particle b.
00:06
Remember we ask questions about when do they collide and where do they collide.
00:12
And we'll take each question as it comes along.
00:15
Before we do that, sometimes it's nice to have a visualization of where these guys are at the beginning.
00:21
And the beginning would be t equals zero.
00:23
So at particle a at t equals zero, put in t equals zero, zero i, and 0j, so it's at the origin.
00:35
Remember, position is displacement from the origin.
00:40
Where are you relative to the origin? that's what position means, displacement from the origin.
00:50
Rb, 0 here, 0 here, but we got a 2 without a t.
00:56
So we get a 6i, then we get 0 minus 2, minus 2, so minus 6j, and put the meters outside.
01:07
Why are they putting the meters outside? because if you were to label all the units that had to be in here on the threes and nines and the twos and all that they will be all different it all depends on the power of t all depends on power t and that would get very very messy very quickly very um cumbersome so you just put what the outside units are that's why we do that so you don't have to deal with all that so b is at six minus six so that'd be down here this is b six and power of x, 6 in the negative y.
01:46
So minus 6.
01:47
6 is y, and x is 6.
01:50
Okay.
01:52
Now, the problem deals with a collision.
01:55
What does that mean? what do we have to look for? same velocity, some acceleration, same exact position, same y position, same x and y position.
02:08
What do we need? well, the last thing i just wrote is what we need.
02:11
They have to be at the same position, x and y, at the same so let me write that out formally.
02:19
R -a, c -quared -r -b, their position vectors have to be the same.
02:24
That would be the same position.
02:26
But remember, as i mentioned, this is a vector equation.
02:29
So that represents x -a equals xb, and y -a -c -o -yb, represents both of those.
02:41
But you've got to be, this must be true at the same time.
02:45
It's possible that at some earlier time, there are some different time that they that b is now at this at x say x a at some earlier time was you know one meter at say one second maybe at uh 1 .5 second xb is at one meter but they're not there at the same time so they're not colliding and that wouldn't be enough anyway as we're going to talk about both have to be true got to be the same x the same y got to be the same position got to be this dot at the same time or maybe not this time i mean wherever it's going to be wherever that collision is going to be got to be that dot that means x and y both so let's xa is equal to xb let's look at that equation first and so we just i'm going to use t sub c represent the time where they collide that simultaneous time same time tc so three tc is equal to three tc squared minus 2 tc plus 2 3s go away so i get tc is equal to tc squared minus 2 tc plus 2 0 is equal to tc squared now when i bring this over i'm going to get a minus 2 tc my tc minus 3 tc plus 2 so there's my there's my quadratic equation now remember, this is the form, a x squared plus bx plus c is equal to zero.
04:32
Solutions of that come from if you can't.
04:35
Sometimes you can factor, you can factor these things out.
04:39
Sometimes you can't.
04:40
And you've got to go to this.
04:42
X equals minus b plus or mi's a square root, b squared, minus 4a, over 2a.
04:52
And that is a, let me fix that.
04:54
That's a c.
05:01
Nowadays, calculators can do it.
05:03
Internet, you can find websites that will do it for you.
05:08
Your computer may have a program on it directly.
05:10
That'll do it.
05:12
Whatever has to be the case.
05:14
But this is where it comes from.
05:16
That's what those programs are built off of.
05:19
If you put all that in, i won't bother you by going through the, take the time of doing the, it's just a matter of punching in.
05:27
You know, your a would be one, your b is minus three, and your c is two.
05:30
It's a matter of putting it in.
05:32
Tc equals one second, two seconds.
05:37
I say, wait, it's going to collide twice? no, no.
05:42
We don't know which one of these is the proper collision.
05:46
Remember, this only tells me that they're at the same x, but maybe, maybe though, a is at this point here, and b is at this point.
06:01
They've got the same x, but they don't have the same y.
06:06
So we've got to now look at ya is equal to yb and find the time that matches one of these.
06:13
We better find one that matches, otherwise we've done something wrong.
06:17
Okay, so putting in our tc here, we get 9 tc, 2 minus tc, is equal to 3, tc minus 2.
06:30
The 9, you can make that into a 3, and this 3 can go away.
06:34
So we get 6 tc minus 3 tc squared is equal to tc minus 2.
06:47
And we can bring all these terms.
06:50
Why do i do it on to the right? because i want to get rid of the minus 3.
06:54
I want to have just a positive.
06:55
It doesn't really matter.
06:57
All comes out the same.
06:58
So 0 is equal to what would i have here? 3 tc squared minus 6 tc plus tc minus 2.
07:08
2, so 0 is equal to 3 tc squared, and we get minus 6 plus 1, so minus 5 tc minus 2.
07:18
There is our second quadratic equation.
07:21
And if you plugged everything in, you get here for when they're at the same y, minus one -third of a second and two seconds.
07:35
So there we have it.
07:37
Two seconds is the collision mark.
07:38
I say, wait a second, what's this minus one -third? what's this minus time? you said it started at zero.
07:43
You can understand, you may have talked about this when you did, when you do just any kinematics.
07:53
The equations are for all time, minus infinity to plus infinity.
07:57
That your actual problem, your actual motion started at t equals zero or whatever time.
08:03
Those equations are, those mathematical equations span all times.
08:09
You can get something in here say, wait a second.
08:11
Nothing was even moving at that time.
08:13
Fine.
08:14
But the equations traced back would have given you that.
08:17
But we ignore it because it's not part of our problem.
08:22
But notice at one second mark, there was like i was saying, at the one second mark, they had the same x, but they did not have the same y.
08:29
Only at two seconds do they have the same x and the same y.
08:33
There's our collision.
08:34
There's our collision.
08:40
So that's how long after they start moving did they collide, two seconds.
08:46
Two seconds after they started to move.
08:52
Okay, now it asks, where did it take place? what position did it take place? well, r -sof -c, i can use either r -a or r -b...