https://youtu.be/xoCr00l48Zs
PHYS 1101: Lecture Eight, Part Six
Okay we have completed step 3 and we have used our trigonometry to complete as many components as we can. Now let’s go down and make a list of our knowns and decide on what variable we need to focus on.
To make this list of our knowns, I find it useful to separate into x and y that way your eyes more easily group all the x variables together while you are looking at the x equations. So let’s go up here and make a quick list. I am going to do x and know that my initial position, x coordinate is zero. V0x = 13.36 m/s. Position, velocity – I am just going to put my time here, my variable is in the middle of the coordinates of both. I don’t know the time.
Okay so that’s at the beginning position velocity. At the end, I don’t know the final x coordinate. Do I know the final velocity component in the x direction? Yes, I concluded that had to be zero if the ship was going to be headed due north.
On the other side, the y variables. The y position initially is zero. The initial vertical component of the velocity is plus 13.36 m/s. That’s position, velocity at the beginning. Position at the end, I don’t know the y coordinate and the final vertical part to the velocity I also don’t know. That’s x, that’s y.
The other thing I can put between the two here is the acceleration components. I know that Ax is -2.075 m/s^2 and Ay is +2.475 m/s^2. Okay given these x variables and the y variables. Let’s see if we can go after the variables that we need.
What do we want? Let’s go back and read the problem. First question – how long does it take to assume the original course again? We need the variable t to answer that. After we get that, then we need to figure out how fast it will be going.
Notice this how fast refers to the final speed – that’s going to be just the magnitude of the final velocity. At the end, at time t and were headed due north in that instance, how long is this vector? The magnitude of this vector is going to be the magnitude of Vy because the x is zero. So we need t and we need Vy. Write that here as want, we want to solve for t and Vy.
Next step, step 5 – to do this step, I went ahead and copied our equations again here to the right so we can look at them more easily. So let’s start here with t, that first question, we want the variable t, here’s a clear list of the x variables we know and of course, we have our set of x equations to use and our y variables that we know and our set of y equations.
Well t shows up in both sets, which is going to help us out. Well my eyes immediately drawn to thinking about the x horizontal part to the motion simply because looking at these lists I know more of these quantities than I do for the y part to the motion.
Can I use this to figure out the time? Well let’s see time shows up in this very first equation. Final horizontal velocity is equal to the initial plus Ax*t. I have Ax, I have the initial horizontal velocity, the final – that equation would work for me, I am going to use that one. I am going to use Vx = V0x + ax where I know all three of those quantities so if I do all my algebra, I subtract that to the other side and divide by Ax. Check my algebra in more detail if you need to. t = Vx – V0x/Ax.
Plug in numbers for that, my final Vx is 0. So I have that t = 0 – 13.36 m/s / Ax which is -2.475 m/s^2. Quick unit check, I am feeling good, the meters cancel in the numerator and dominator. One of those seconds cancels with one of those seconds. The second in this dominator in the dominator is going to go up to the numerator.
Also notice that by carefully keeping track of the signs, the correct sign values here ended up in a negative and a negative cancelling so time is positive which is good. That’s what you would physically, physically make sense. I end up with t = 13.36/2.47 that is 5.398. Keeping four digits that’s a little over 5 seconds, I would actually type into WebAssign 3 digits and I would round to that 5.398, the 8 would round this up and that would make this 40.
The next thing we needed to know was the final speed, the final speed the magnitude of the final velocity. That’s how fast it would be going at that instance. And you remember our final velocity was a vector pointing straight up which means it only has a y component so what we really need to focus on is the variable Vy as I said here.
So let’s look at our y equations and our list of y variables and see what we know, what will get it for us. Let’s look at this first equation, that’s looking really good to me. I want Vy, I do know what the initial vertical velocity is, I know Ay, and I know now t. I can use that equation directly. T is now a known value. If it took long for the horizontal velocity to go back to zero and in that same time, I can calculate what happened to the vertical part of the velocity.
So I am going to use equation one of the y equations. Vy = v0y + Ay*t. I can just plug in numbers directly for that. V0y was +13.36 m/s plus the vertical component of my acceleration was +2.475 * time and my time is – I’m going to plug in the full four digits for that because then at the end for this final answer, I am just going to grab the three digits. I am going to put in 5.398 s.
Let’s do a quick unit check. This second cancels with one of those so I am left with m/s. That does make sense to add to a quantity that is m/s and of course, that’s the units I am looking for.
When I plug this into my calculator, I get 26.7 m/s rounding to my three digits. So my two answers are it takes 5.40 seconds for that boat to get going due north again and after it’s done that, its headed due north with a y component to its velocity or its velocity vector having a magnitude of 26.7 up – that is the speed, that is how fast it is going so that is what I would type in.
So you notice this is quite a bit bigger than the initial velocity that I had. Does that make sense? Quick let me go back to our picture here, let’s see if it makes sense. My acceleration in this direction, it does have this negative x component to it, so that’s going to decrease the horizontal velocity and bring it to zero. At the same time it has a vertical acceleration component here – that’s going to cause the vertical part of this velocity during the same time interval to just keep getting larger and larger by that delta-v. So I probably didn’t exaggerate it enough, that this final velocity is actually quite a bit bigger. It is roughly double than what this initial height was. That’s a little overboard now. It looks like that.
So that’s a good picture of velocities at the two snapshots giving this acceleration of what our ship does to this problem. So for this last point here, the last step to think through and make sure your numbers make sense, I went back and looked at the acceleration vectors and velocities, checked our units along the way, as best as we could tell it does look all and good. It does make sense.