Vectors and Projectiles Legacy Problem #11 Guided Solution

If I were to practice the habits of an effective problem solver on this problem, I would read it carefully, and then I would construct a diagram of the physical situation. And as I read this problem, what I determine is that there's two planes. One of them is heading north, and the other is heading south. So I'm going to begin by putting a dot for that northbound plane on a sheet of paper, and then the other plane is north and west of that dot. In fact, I'm told that it's 102 degrees from that dot. So if I think 102 degrees, that's a vector that goes 12 degrees west to north. So I think of a line drawn from my first drawn dot heading due north, and 12 degrees west of that line is our other plane. So I put a second plane a little bit off to the west of that northerly line, and it's 13.5 kilometers away. Now I could label that as plane 2, or I could put an S next to it because it's traveling south. My first dot that's on the southern side is the N dot. It's heading north. I could put little arrows there if I wish to indicate the direction of motion. But what I'm asked to find is how many miles north and how many miles west is the second plane from the first plane. In order to calculate that, I need to use SOH CAH TOA, or some variation of that. Now, one of the big ideas that we have in this unit is that we can find the X component of a position displacement velocity acceleration force vector. If we take the magnitude of the vector, in this case 13.5 kilometers, and multiply it by the cosine or the sine of the angle that that vector makes counterclockwise from due east. So applied here in this situation, I can find the northern component if I go 13.5 times the sine of 102 degrees. And I can find the western component if I go 13.5 kilometers times the cosine of 102 degrees. This gives me the north and the west component of this displacement from first plane to second plane. And once I've done that, I can prepare to solve for part C of the problem. In part C, I have two planes heading at 290 kilometers per hour, each of them heading at 290 kilometers per hour. And I've just figured out the north-south distance between the two planes. That's part A of this question. Now, if I want to find the time it takes before the two planes are side by side, well, I'm going to have to use an equation that has a d and a v in it and a t in it. And the equation is simply d equals v times t. Normally, we think of a 1 half a t squared term being in this equation, but it drops out since there's no acceleration being spoken of. So the distance traveled by either one of the planes is simply half of that distance that you calculated in part A. Half because each plane is heading towards each other, and they meet at the halfway point. So if I want to find the distance that either one of the two planes travels, I take that separation distance, the answer to part A, and I cut it in half. And I put that in for d, and then I use my v, 290 kilometers per hour, and I solve for t. And that's how you solve part C of this problem.