 It's a beautiful day, so let's take a break from physics and go for a paddle in the river. Here's a view of the river from above, and let's say we can paddle at two meters per second in any direction we like. But to make things interesting, let's have that river start moving with a current of one meter per second to the west. So you probably know that if I paddle in the same direction as the river, my overall or resultant velocity will be three meters per second, and if I paddle in the opposite direction of the river, my resultant will be only one meter per second. So I guess I go faster when the river is with me and slower when I'm against it. That makes sense. Now let's say I want to get from point A on the southern bank of this river to point B on the northern bank. So what's going to happen when we aim our canoe north towards point B? Well, the velocity of the river current is going to push us off course, and we won't go exactly north. We'll go north and a little bit west. We can see our overall velocity, what we call our resultant velocity, by drawing a proper vector diagram. We'll draw the velocity of the canoe and the velocity of the river as the components, so they'll go tip to tail, and we'll draw the resultant velocity in the direction that will actually go tip to tip and tail to tail to the components. We can even work out how fast we'll be moving. We can use the Pythagorean formula to get the magnitude of the resultant, which works out to 2.2 meters per second, and the inverse tan ratio to get the angle, which will be 27 degrees west of north. So how far down river did we end up? For this, we need the width of the river. Let's say it takes 10 meters to travel from point A to point B. We can add that vertical, or y-displacement vector to our diagram. Here's the vector that represents the displacement of the canoe going downstream, the horizontal, or x-displacement. So how are we going to figure that dx out? Well, we can model the motion of the canoe with a uniform motion formula, v equals d over t. This makes sense since the canoe is not accelerating while it's in the river. Substituting into this formula is a lot trickier, though. We can put in 10 meters for the displacement, I guess, but what velocity should go in? There are three different choices here. There's a key idea here that helps you work this out. When you're working with kinematics formulas, you need to substitute in x-vectors with other x-vectors and y-vectors with other y-vectors, or, as I often say in class, keep your x with your x and your y with your y. This means that if you substitute in a y-displacement, like 10 meters, you have to also sub in a y-velocity, in this case 2 meters per second. That will allow us to calculate a time of 5 seconds, which is how long the canoe is in the water for. After we know the time, we can do a second uniform motion calculation for the x-direction. This time, we'll use the x-velocity of 1 meter per second, along with the time to get the x-displacement of 5 meters. Lastly, what if we wanted to go from point A to point B exactly, without drifting any? How could we make that happen? Well we would need to paddle our canoe into the current, actually aim ourselves a bit to the east to counteract the motion of the water moving to the west. We can draw that out with another vector diagram. This time, the velocity of the canoe is the diagonal vector, and is aimed a bit to the east into the current. These two vectors are still components, and they're still tip to tail. Even though it's kind of weird that the green canoe velocity component ends up being the diagonal. The resultant, our overall direction of motion, will still be tip to tip and tail to tail with the other two components, and will point straight north. Note this is an example of a resultant not being the hypotenuse of the triangle. We can even calculate the magnitude and direction of this resultant. Just be careful when you use the Pythagorean formula that you make sides see the hypotenuse of the triangle, two meters per second, and then solve for either side A or B, which is the resultant.