00:02
Okay, so in this problem, we have lucas, who lives by nearby a calm stretch of cowican river, and the speed of the current is around one kilometers per hour, so i drew that there.
00:16
I also drew the direction of the downstream here, and he basically goes on a trip.
00:21
He goes five kilometers downstream, and then turns around and goes six kilometers upstream to see his grandfather, so he starts here and then he ends like one kilometer to the west, i guess the upstream from where he started.
00:39
And i'm assuming that he's been canoeing at the same speed entire time.
00:43
That's why i've had the boat here.
00:45
In either direction, there's v.
00:47
V stands for his speed.
00:49
Not velocity.
00:51
We're just using speed here.
00:53
And then we're asking, well, okay, what would lucas canoes speed be on still waters? okay, so whenever he's going downstream, maybe i could just write a table here.
01:09
So whenever he's going downstream, what would be his speed? well, his speed is going to be v, his usual speed on calm waters, plus one because he's aided by the downstream.
01:28
Upstream, it's the opposite.
01:33
He has his usual speed, but then he's kind of, being struggling against the water the downstream here now how much time would it have taken him well we don't really know but we know the total time is a whole three hours so we can just say that the time downstream was t and if so then the time upstream would have been three hours minus t okay now what about distance well we know that he went down five kilometers and went up six kilometers.
02:17
Okay, so with these pieces of formation, we can actually solve this, we can actually solve the question here, mainly because of this fact.
02:27
Distance equals speed times time.
02:38
So what does that mean? well, we have five for distance in downstream equals v plus one for speed times time.
02:48
Which is t.
02:50
On the other hand, we have six upstream, v minus one times three minus t.
02:57
And voila, we have a system of two equations and two unknowns.
03:02
And eventually, with some algebraic manipulation here, we should be able to solve it.
03:08
Now, how would we go about doing it? i kind of just experimented.
03:12
You can just play around with it, honestly.
03:13
Or you could make your matrices and whatnot.
03:17
It's more fun.
03:18
It's like plato.
03:19
Just have fun with it.
03:21
So vt plus t.
03:22
I just expanded both of them.
03:25
And if i expand both of them here, the second one, don't forget your foil, or in other words, just basically pair up every term in every parentheses with every other term in every parentheses.
03:40
So we get that.
03:43
So what do we get? we get 3v minus vt, minus 3 plus t.
03:53
Oh, look at that.
03:54
This is a vt right here.
03:56
Maybe i should highlight it.
03:59
Yes, i should highlight it.
04:01
We have a vt here and then negative vt here.
04:05
So if we can just add these two equations together, we could probably get away with something.
04:11
So if we add, or rather, keep the first equation, but then second equation, you change by adding the first equation to your second equation.
04:21
So we get 11, 5 plus 6 is 11 equals.
04:24
And then if we add these two guys together, they get canceled out...