https://youtu.be/m7kYB_8Ne1k
PHYS 1101: Lecture Eight, Part Five
Okay, let’s do an example. Our problem is that we have a shipping moving, initially at 18.9 meters per second due north. This ship drifts due north east at the same speed. To correct the bearing, the captain accelerates the engine and turns the rudder NW. If the engine delivers a constant acceleration of 3.5 meters per second squared how long will it take to assume the original course again? And how fast will it then be going?
What I’ve got pasted below here, are the explicit problem-solving steps. Of course, the point to these steps is all about helping us understand what these variables are in these equations that we are going to leverage or use to solve the problem. To determine how long it is going to take to resume the original course, and then how fast it is going to be going.
Step one is visualize what is the object, what is the scope of the problem. The start and the end. Picture a motion diagram, or at least a trajectory of what this motion is, initial velocity, final velocity, and acceleration.
The object, I think is clear, it’s the ship, whose motion we are discussing. The start and the end are a little subtle in this problem. Remember that the scope of our problem has to be between two instances during which the acceleration is constant.
In this problem, the engine delivers this constant acceleration, as I read the problem, between the instant where initially, the ship is drifting northeast at 18.9 meters per second. That is the initial snapshot.
Then the final snapshot is our goal of the instant where the boat then resumes its original course. Or the final instant is, “How long does it take for the boat to be moving again, due north?”
I’m going to jot that down so I have it in my mind. “The start is actually when this boat is drifting”. I’m going to grab a green marker, here, to just sketch to myself.
Off to the side, I’m going to do my north, south, east, west, for orientation. We’re going to call these the “positive directions” when we map it on to our x and our y.
Given this orientation, direction, my initial velocity vector is pointing northeast. Northeast would be 45 degrees, like this. And the initial speed, which would be the magnitude of this initial velocity, is 18.9 meters per second.
This is the initial trajectory of this ship. Eventually, at the end of the problem, this boat has to be going, again, due north. Its velocity vector then, at the end, would be pointing straight up, due north, because then it will have resumed the direction in which it was originally going.
Between this start – maybe I should jot that down, here – between this start and this end, the constant acceleration I have is this 3.5 meters per second squared. What direction is this acceleration vector?
The direction information has to be coming from this section of the problem. The captain accelerates the engine and turns the rudder northwest. This I take to be the direction of this acceleration vector that has a magnitude of 3.5 meters per second squared.
Let me draw a red arrow on my sketch, that is northwest, and just jot down, that must be a that I am after. This helps me picture what is going on.
With this constant a, this constant little delta v, it’s going to take this initial velocity vector off in the northeast direction, and gradually tilt it and change it, to swing the boat around and have it head due north.
The trajectory, then, if I were just to sketch it off to the side, would look something like this, for the total path that the object followed. This is the velocity at the start, and this is my straight-arrow velocity, right at the end.
That is all I would do for step one, here. That has given me enough to clearly visualize the scope and these vectors. The next point is to draw my axes, define origins, and then to start thinking specifically about what these variables are.
In order to do that, I’m going to go up here. I didn’t realize it was so far away. To do that, I’ve gone up to a previous point in our lecture and copied our generic axis. Let me make this smaller.
Because this, in essence, is what I want to draw or create for this particular problem. Rather than this generic one that I used to introduce the idea to you.
What does this plot look like for our problem? I would propose, and I think this is, in general, always possible to do, and is useful, is to put the object at the start of the problem right at the origin.
This will be nice, because then your initial positions are going to be 0 and 0, for both x and y. Here is t0, the object, the ship, is going to be right at the origin. At that instant, I know that my initial velocity points northeast, so I’m going to draw that arrow, right at that instant, to remind myself of that.
Now the acceleration vector for this problem is not down and to the right, but rather, is pointing due northwest. So I’m going to copy that over here, and just draw it somewhere near my problem.
The endpoint for this problem is with the ship heading due north. It accelerates from the start at this 3.5 meters per second squared, which is what curves this velocity around, until it is going due north.
For the end instant in time, I’ve concluded that a more realistic location for that would be something like this. That the object keeps moving up, so I’m going to sketch that as my final location at time, t. It will have the final coordinates x and y. And I know that my final velocity is up, at that instant.
I’ve moved that down for us, a little bit. Let me erase this final velocity and jot down that for this problem, the final velocity is up, due north. Let me go ahead and grab my acceleration and move it somewhere near this region a bit. And I’ll move my y axis, just stretch it up a bit, here.
There I have slid it over and tidied it up again. But I’m showing you really clearly here, the key vectors and the picture that you need to have, as we start to break down these vectors and identify the values of the various components.
Here is my initial position. I’m going to put my origin there. Finally, the ship ends up and to the north. It ends up northeast, at some time later, after accelerating with a magnitude of 3.5 meters per second squared, in this direction.
That is the result of the final position.
Okay, for how many of these variables do we know values for? t-0, we always start at 0. The initial position, I put it at the origin, so both the x coordinate and the y coordinate are 0, right here, initially.
The initial velocity components, I can get those doing Trig, and I’m going to do that in just a minute. The final velocity components, I don’t know, but I do know the direction of this vector. It is pointing due north, so in fact, that does tell me that, at this final time, at this final location.
I may not know what v sub y is. v sub y would be the total final velocity vector. It only has a y component. but I do know that v sub x has to be 0. And of course, my components for the acceleration I can get as soon as I do a little trigonometry on this number.
Here I have just drawn them over on to my drawing. Sometimes it is helpful to sketch them there, where all of your initial variables, time, position, velocity at the start, somewhat line up with that location on your drawing. Then at the end here I’ve got time, position, and velocity.
Let’s go back up and read the problem, real quick, and be sure that we’ve gotten all of our initial information. Yeah, all I know is that I can get my initial velocity components based on the fact that the speed, or the magnitude is 18.9.
Then I have the magnitude of the acceleration, but there really is no other information. I don’t know final positions. I don’t know the total amount of time. So that’s all I have for step two.
Before we officially list all of the values, the things that we know, let’s do some trigonometry on as many vectors as we can. Look at your vectors. You’ll really have just three of them: v0, a, and v.
Of these three, which can you do trigonometry on? What do you know? My initial velocity vector had a speed of 18.9 meters per second. That is hypotenuse, 18.9. I need to find the two sides of that triangle. That will be my v0y, and my v0x.
This angle, I know, is 45 degrees. v0x, this side of the triangle, is going to be 18.9 meters per second times the cosine of 45. That gives me 13.36 meters per second.
The direction of that, the velocity vector initially, is up and to the right, so vx has to be to the right, and that is going to be positive. I’ll leave it positive. v0y is going to be the hypotenuse times the sine of 45. That also is 13.36 meters per second. It is also going to be positive, because v0y captures the upward trend of that result vector.
Okay, there is a vector for which we can find the components. The final velocity vector, as we stated above, the only thing we can conclude about that is that vx is 0.
Here is our third vector, the acceleration. We also can do some trigonometry on that to determine its components. Let me draw that vector as a right triangle. I’ll expand it a bit.
a points in the northwest direction. So ax is going to end up being negative, and ay is going to be positive. The hypotenuse is 3.5 meters per second, so ax is going to be negative 3.5 meters per second. This again, is 45 degrees times the cosine of 45 degrees. That gives me minus 2.475. I’m going to keep one more unit.
For Web Assign, keep four digits as you work the problem and round to three at the end. That will minimize your rounding-error problems.
ay, the vertical component, is positive, because it points up. But it’s going to have the same magnitude. When you take 3.5 meters per second squared times the sine of 45, you’re going to get the same magnitude.