Vectors and Projectiles Legacy Problem #20 Guided Solution

This problem falls into a class of problems known as a riverboat problem. Before I actually discuss the solution, I'd like to call your attention to a link that you see on this page, to a page at the Physics Classroom tutorial. That page at the tutorial is called the Relative Velocity and Riverboat Problems page. It takes a very systematic approach to discussing how to solve such problems, giving you step-by-step directions, and also giving you much depth in the way of diagramming and pictures. It's a very good read. OK, now for this problem. What we have is Tyron Lakes is paddling across a very lazy river. The river is flowing south, which tells us that the banks of the river run from north to south. So I begin by sort of drawing a diagram, and on the diagram I draw the banks of the river, and then I notice Ty is heading west, so I go to the rightmost side of the river and I draw a vector towards the west or towards the left, and I label that as Vboat equal 98 cm per second. That's the velocity of Ty's boat. Now, if there were no flow of the water, Ty would end up directly opposite where he started on this river, but instead there's a flow of the river southward, which carries Ty's boat southward. So I'm going to draw that southward river flow, another little vector a little shorter than the first one I draw, from the arrowhead of the westward vector, and I draw it south. I label it Vriver equal 48 cm per second. Now, what I wish to do in part A is determine the magnitude and direction of the resultant velocity. That is, what's the sum of these two velocities? I have one being 98, the other being 48, and I add them together to get a hypotenuse of a right triangle that's heading a little bit south and a little bit west. I draw it on my diagram, and I find its magnitude using the Pythagorean Theorem, and when I do that, I get as a value for the magnitude of that vector, I get 109 cm per second. And for the direction of the vector, I have to go back up to the tail of that diagonal resultant, and I have to find the angle that it makes with west. So I go inverse tangent a side opposite the side adjacent. So it would be inverse tangent of 48 divided by 98, and when I do the math on that, I get about 26.1 degrees, and that's the number of degrees south of west. If I had drawn my triangle a little bit differently and gone to the tail of my resultant vector, then what I might have been doing is measuring the angle 64 degrees west of south. It doesn't matter how you express it, that's going to be the same direction either way. Now, once you get the magnitude and direction of the resultant velocity, you go to Part B, where they ask you to find the time to cross the river. Now Tai is paddling his boat west, the river is pulling him south. What you're given in terms of a distance is you're given an across-the-river distance, 22 meters wide. That's from the east bank to the west bank. That's a horizontal distance, and since you have a horizontal distance, to find the time to go that horizontal distance, you should use a horizontal velocity. That is, you should use 98 centimeters per second. Now right away you might notice a problem here. We've got meters and we have centimeters per second. So you need to do a conversion from either meters to centimeters or centimeters to meters so that you're not mixing up your apples and oranges. Then you're going to use the equation d equals v times t. Well, I've chosen to use, as my unit, meters. So I'm going to go 22 meters for d equal vt, where the v is 0.98 meters per second. And when I do my math on that, I'm going to get a time in seconds, and it comes out to be 22.449 seconds, and I can round that to 22 seconds. Now, in the next part of my problem, I'm going to try to find the distance traveled south, that is, downstream. What carries you downstream is this 48 centimeters per second reverse speed. Now, that's centimeters once more, and I wish to find my answers in meters, so I'm going to convert that to 0.48 meters per second. And again, I'm going to use the equation d equals vt, but this time to find a southerly distance, I use 0.48 meters per second, and that time from part b, which was 22.449 seconds. When I multiply them together, I get about 10.7755 blah blah blah meters. I can round that to 11 meters, and that's the distance traveled south, or downstream.