4. 2D Kinematics
Kinematics in 2D
Kinematics in 2D
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Hey, guys. So now that we've covered all the motion variables in two dimensions, you're gonna run into problems which you have to solve Kinnah Matics problems in two dimensions. So I want to show you this video is that solving problems with constant acceleration in two dimensions is done the exact same way that we did this for one dimensional motion. We're gonna use the same list of steps with just minor differences and the same list of equations here to solve these kinds of problems because remember that the whole point of 22 dimensional motion is that we could always break it up into 21 dimensional emotions and then tackle them separately. So we have everything we need. We just jump in and solve this problem. So we got a hockey puck that is sliding along the lake with 8 m per second east. There's a strong wind that accelerates it. 3 m per second, some direction. We're gonna calculate the magnitude and direction of the displacement after five seconds. So the so the first step is always we're gonna draw the diagram Now. The only new thing is we're gonna have to decompose any two dimensional vectors into X and y. So let's check it out. So we've got a hockey puck that is sliding along the east. So this is my V not. And we know this is 8 m per second. Now there's an acceleration that is that some angle 37 degrees northeast. So we've got a acceleration vector A equals three, and we know that this angle here is 37 degrees. So now we're gonna figure out what the displacement after five seconds, what happens is they're they're the acceleration in that Northeast direction is gonna start curving the hockey puck. It's gonna start curving it like this. So eventually, at some point after five seconds, we're gonna try to figure out the magnitude of the displaced, another direction of that displacement. So for starting from here than the magnitude, it's gonna be the length of this high pot news. This is the magnitude Delta are, and the direction is going to be given by the angle relative to the access. Right. So now we have a bunch of two dimensional vectors. We're gonna decompose them all into X and y. So what I'm gonna do is this is my displacement vector like this. So I'm gonna call this Delta Y I've gotten accelerate and then Thean X direction here is gonna be Delta X And then I do the same thing for my a vector, right? So I've got a component in the Y direction like this a y and a component in the X direction like this, a x alright. And then my velocity vector is already in the extraction, so I don't have to do anything to that. All right, so that's the first step. Now I'm going to list the five variables for Delta X for X and Y I'm gonna identify known in target variables. But first, let's take a look at what I'm actually asked for. The problem. A mask for the magnitude and the direction which is the Delta r and D theta Delta are right. So let's take a look at my equations. I know the magnitude for Delta R is just gonna be the Pythagorean theorem Delta X squared plus Delta y squared and the direction theta Delta R is gonna be the tangent inverse of my delta y over Delta X with the absolute values. So notice how both of these equations actually involve Delta y and Delta X. So I'm actually gonna have to solve for those first before I can. I can actually solve for the displacement, um, for the magnitude and the direction of the displacement. So let's move on to the second step. Now I'm gonna have to list the five variables for X and y, so I'm gonna have my ex direction here, my Y direction here. Remember, there's five variables a Delta X V not in the X Direction V find on the X direction A X and then T and the same thing for why dealt why? I've got V initial in y and then v final and then a y and anti. So we're gonna have to identify the known variables and our target variables. Well, we know that in order to solve for the magnitude and the direction, I'm actually gonna need Delta, X and delta y to basically put them into the two dimensional displacement vectors or equations. So these are my target variables. So what about the knots? What can I figure out the initial velocity? Well, I'm told that the initial velocity of the puck is 8 m per second, and it lies purely in the east direction. So what that means is that this vector over here, all of it lies along the X direction. So v not in the X component is eight. So I know what this is. This is eight. And because all that lies purely in the east direction, that means the initial velocity in the Y direction is zero. There is no component that points north, if you will. So this is gonna be zero. All right, So what about final velocity? Don't know anything about final velocity. What about the acceleration? I'm told the magnitude and direction, but I don't have the components. So let's move on to time. Well, I know that the time is equal to five seconds, and it's the same for both. You don't have to, you know, distinguished X and y. So basically, what happens is I have two out of my five variables in the X and y, which is not enough. I'm gonna need one more of these equations in either X and y to pick an equation. So between in my final velocity and my acceleration vector, let's take a look. Which one I could solve. I'm not told anything about the final velocity, but I do have the magnitude and the direction of the acceleration vector. So I have a and I have theta, which means that I can use vector equations to solve for X and A y. So my a X So let's go over here. My A X components is just gonna be a times the cosine of data. So it's gonna be magnitude, which is three times the cosine off 37 degrees. And so what I get is I get 2.4 and I do the same thing for the y. I'm gonna get three times the sign of 37 that's 1.8. So basically, I know this A X is 2.4, and this a Y is 1.8. So now I actually have a X anyway. All right, so now I have three out of my five variables, which means I can move on to the next step so I can just pick, um equations without the ignored variable. So in this case, my ignored variable is gonna be my final velocity. I don't know anything about it and I don't care about it. So this is my ignored variable and I'm trying to solve for Delta, X and Delta y so in the X direction, I'm gonna use equation number three, which is that right? Because this one doesn't have the final velocity. So I'm gonna use that Delta X is equal to V not x t plus one half a x t squared. So I know that the initial velocity is eight. The time is five plus one half, 2.4 times five squared. You'll get 120. So that is my Delta X. And so if I do the same the next thing for the Y direction I'm gonna use equation number three because I have the same exact unknowns. So my delta y is gonna be V not y t plus one half A Y t squared. So this is how is gonna be zero because my initial velocity is zero. And then this is gonna be delta y equals one half of zero of 1.8 times five squared, and I get 22.5. This is meters. This also meters. So I have my displacement in the X and y directions. Remember, I needed to get them to plug them all the way back into my Pythagorean theorem and my tangent. Inverse. So for the final answer, my magnitude is gonna be Pythagorean theorem of 120 squared plus 22.5 squared, and I get 122 meters. And if I do the data, my tangent adverse. I get 22.5 over 120 this gives me an angle of 10.6 degrees. So these are my answers. And that's really it for this one. So let me know if you guys have any questions.
A survey drone has just completed a scan at x,y coordinates (57m, 8m) at t=0. It needs to return to a lab located at (-115, 72) m. If its initial velocity is 16m/s in the +y-direction, and it has only 18s of battery life remaining, what constant acceleration (magnitude and direction) does it need to reach the lab?
2.8 m/s2; along –x axis
1.8 m/s2; 51.8° below –x axis
2.8 m/s2; above –x axis
1.3 m/s2; 24° above –x axis
Additional resources for Kinematics in 2D
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