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Motion Along a Line - Problem 3 1,816 views
We're talking about motion along a line. In this problem, I have a red ant, and a black ant crawling so that their positions are given by these two vector equations. The first part says, sketch their paths. Do their paths cross? So I want to sketch these two vector equations.
Now I've got the vector equations written up here. I'll first sketch the top one, that's the red ant. 19,1 this is going to be the initial point. 19,1 is 5, 10, 15, 20 is here so 19 is here, 19 would be this point. The direction vector is -2,1. So I could get a second point by going 2 to the left, and up 1, but I think I'd rather go a little bit further. So let me go 20 to the left to -1, and up 10. So 5, 10, and here will be a second point to use. Let me graph that point.
It's better to graph points that are far apart when you're graphing lines, you get a nicer line. So that's the red ants path. The second equation starts at -1,-7. So -1 is here, -7 is down here. So that's the starting point, and then we need to find another point using the direction vector 3, 4. So I can go 3 to the right, and 4 up. 3 to the right, 4 up. 3 to the right, 4 up, you get another point, and then I can draw my line.
So you can see that their paths do actually cross, but does that means that the two actually meet? That's what we're going top answer in the next question, but let's think about that right now. Just because their paths cross, does not mean that they actually meet. For example, at the time when one of the ants is at this point, the second ant might be down here. So we don't know that they're actually ever going to meet. They can be travelling something like this.
Let's find out. So, do the ants ever meet? If so when? Do they paths cross? Yes, they do. The first thing I'm going to do, is write parametric equation for each of these guys. Now first, notice for the red ant, x equals 19 minus 2t. Y equals 1 plus t. For the black ant, x equals -1 plus 3t so 3, and y equals -7 plus 4t.
Now the way I'm going to approach this is, I'm going to set these two guys equal to each other. Because if the ants ever do meet, they'll be at the same x and y coordinate at the same time. Let me start by setting the x coordinates equal to each other, and solving. So 19 minus 2t equals -1 plus 3t. I solve this by adding 2t to both sides, and adding 1 to both sides. I get 20 equals 5t. So t equals 4.
Now all this means is that at t equals 4, they both have the same x coordinate. Let's find out what that is. At t equals 4, the red ant is at 19 minus 2 times 4, 19 minus 8. 19 minus 8 is 11. The black ant is at -1 plus 3 times 4, -1 plus 12, also 11. They're not necessarily at the same position, the y values could be different, so let's check that.
For the red ant, the y value is 1 plus 4 5, and the black ant y equals -7 plus 4 times 4, -7 plus 16 9. So they do not meet. At the time when they have the same x coordinate of 11, the red ant is at y equals 5, and the black ant is at equals 9. So no, they do not meet.
Let's just take another look at the graph, so where is x equals 11? Here is 5, 10, 11 is right here. So at t equals 4, the red ant is right here, at 5. The black ant is up here at 9. So their paths cross, but never actually meet each other.