Vectors and Projectiles Legacy Problem #8 Guided Solution

A relatively common physics lab involves going into the hallway with the students, using meter sticks and finding a starting location, and then tracing out a collection of movements through the hallways, north-south and east-west, to a final destination. These measurements for each leg of the trip are made, and the resultant displacement from the starting location to the finish location is typically measured. That's the basis of this problem with Mackintosh walking from the classroom door, the starting location, through the hallways to some final location in order to determine the magnitude and direction of the resultant. I wouldn't think of drawing this problem without some form of organization or map or diagram or something like that. My method here is going to be to focus first on all the east-west displacements. There are two of them, 32.5 meters east and 68.5 meters west. I'm going to take those two displacements and add them together as vectors, and what I get when I do is 36 meters to the west. So one of the things that happens in this trip is Mackintosh displace themselves in the classroom door, 36.0 meters towards the west. But there's also a north-south displacement, so I'm going to take the three north-south vectors and add them up as vectors. That's 40.0 meters north, 15.5 meters south, and 2.5 meters north. When I add those up, I get a total north-south displacement of 27.0 meters. Now I'm going to take the two lakes in this trip, 36 meters west and 27 meters north, and add them together as vectors. I'm going to do this with a diagram. So I'm going to draw a vector off to the left on my page of paper. That would be west on a map, 36.0 meters. Where that one ends, I'm going to draw another vector north from the arrowhead of that first vector. It's going to go a little bit further, a little bit less far to the north, 27.0 meters. Where it ends, I'm going to put an arrowhead, and then I'm going to draw a resultant vector from the starting point to the finish point. From the tail of the first vector I drew to the arrowhead of the last vector. That's a vector that's heading a little bit north and a little bit west. And I have to find its magnitude and direction. I recognize that that is the hypotenuse of a right triangle, and if I know the sides, I can find the length of the hypotenuse, and I do. So I go 36 squared plus 27 squared equal the square of the hypotenuse, and I solve for the hypotenuse magnitude. Finally, I have to find the direction of this vector. It's not enough just to say it goes north or northwest or a little north or west. I have to find how many degrees north or west it goes. So I look at my resultant, which is the diagonal on my diagram, the hypotenuse of the right triangle. And I notice it makes some form of an angle with west. And I ought to be able to find that angle, because it's an angle inside of my right triangle. I ought to be able to find it using the tangent function, because if you think of that angle, the tangent of its value is equal to the opposite side, 27, divided by the adjacent side, 36. So I say theta equal the inverse tangent of 27 divided by 36, and I find out what my angle is. That happens to be the direction that that vector makes with west, and so I would say it's so many degrees north of the westerly direction, or north of west. I could also find the direction that this vector makes counterclockwise from east. I imagine the vector that I've drawn, that I'm focusing on, that's going both north and west, has got to have an angle counterclockwise from east that's more than 90 degrees and less than 180 degrees. In fact, it's less than 180 degrees by the amount that I've just calculated. So I'm going to subtract this theta value from 180 degrees to get a number of degrees rotation counterclockwise from east.