Vectors and Projectiles Legacy Problem #7 Guided Solution

An accurate solution to a problem such as this one is going to require a carefully constructed diagram in which you represent the motion of Sheila and the cross-country team from the school to the park. Sketch yourself a sheet of paper or you can make a sketch and give yourself some room. Put a little dot somewhere on that paper and call that dot the start location. And then from the start location, begin to trace out the vectors for this cross-country team. The first vector is a northerly vector that goes .68 miles north. Just draw a little northern vector, put an arrowhead on it, put a dot at the end and where that dot is, you're going to start your next vector. It goes off to the right or east on a sheet of paper, east on a map to the right on a sheet of paper, 1.09 miles. It's longer than the first one. You can go ahead and draw it to the right and then put an arrowhead at the end of it where that arrowhead ends, begin to start your third vector, which again is going to go up on your sheet of paper, north on a map, 1.56 miles. That's longer than the first one and longer than the second one. Put a dot there and then prepare to draw your fourth vector, which is .32 miles to the west. Now that .32 miles to the west is taking us backwards horizontally towards the initial position, but horizontally, simply west. Draw the vector. It's shorter than all of the other three. Put an arrowhead on the end of it and call it .32 miles. Now you can draw the resultant because the problem is about finding the magnitude and direction of the resultant. And the resultant goes from the start to the finish, the tail of the first vector to the arrowhead of the last vector. Now that's a diagonal line that's going a little north and a little east. I should say a little east and a lot north. Now we can find out how far east and how far north that is if we think in terms of vectors. We have two northerly vectors here, the .68 and the 1.56 miles. And that gives us the total north movement. You can add those two quantities together and that forms one leg of a triangle that has as a hypotenuse the resultant vector that we've just drawn. And then the east-west side of that triangle is found from taking the 1.09 miles east and subtracting the .32 miles from it because that was west. And that forms together, the 1.09 minus the .32 miles, forms the east-west leg of that resultant vector. If you wish, you can go ahead and redraw that vector and give its sides as .77 miles, which you get when you take the 1.09 miles east and subtract .32 miles from it. And as the other side, 2.24 miles, which is what you get when you take 1.56 miles and you add .68 miles to it. Now you have a right triangle with sides of .77 miles and 2.24 miles. You can find the resultant's magnitude by applying Pythagorean theorem. The .77 squared plus the 2.24 squared is equal to resultant squared. Solve that for the resultant. Then take a deep breath because the harder part of the problem is trying to figure out the direction of this displacement. Well clearly this is a first quadrant displacement that's going north and going east. And we can find the magnitude of the direction, the value of the direction, if we use a sine cosine or tangent function. I typically pick tangent in situations like this and I say the tangent at theta is the side opposite to the side adjacent. Now you might be asking, what angle are you talking about? We're talking about the angle that that hypotenuse makes with west. In other words, it's the bottom left corner of that right triangle that you redraw. So I go inverse tangent equals 2.24, that's the opposite side, over .77, that's the adjacent side, and I find out what that is. And I get my angle. It turns out to be about 71.0 degrees, and that's 71.0 degrees north of east. Or you could say counterclockwise from east.