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concept

## Acceleration in 2D

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Hey guys, you may be asked to calculate the acceleration of an object that's moving in two dimensions. So not in the X or why, but at an angle like this. What we're gonna see is that it works very similar to how velocity in two dimensions works. We're gonna end up with a very similar set of two equations to jump back and forth between the acceleration and its components. These equations are gonna look almost identical. It's really only just the letter that's different. So let's just talk about the differences and then do a quick example. Alright, guys, remember that acceleration always causes a change in objects velocity. It changes either the magnitude or the direction of the velocity or sometimes even both. And this actually works for one dimensional motion or two dimensional motion at an angle. So the equation that we use was a equals Delta V over Delta T. It's gonna work the exact same way in two dimensions. It's just that now that we have things that angles, all right. So for for the velocity vector in two dimensions, we had two different equations to calculate the magnitude. We they were given the displacement. In time, we calculate V or we could use the components of the vector so we would take this two dimensional vector and break it up into its components VX and vy y. Now, up until now, we've always visualized these sort of magnitudes directions and components as triangles because they help us visualize all the triangle equation that we're gonna use. Well, from now on, we're actually gonna start drawing all the components starting from the same point like this. So this is the view I component. The math is the same. It's really just only the notation that's different. And so the high pot news of the magnitude, it's just the Pythagorean theorem v x squared plus view I square. The same exact thing works for acceleration. So you have two different equations. You could either always relate this to the change in velocity of a change in time. So if you're giving Delta V and Delta T, you can calculate the magnitude of a or if you're giving the components a X and a Y, then you can calculate the high partners through the triangle by using the Pythagorean Theorem X squared plus a Y square All right, let's move on. So the angle of this vector here is just related to the components by the tangent universe equation. Right? Three angles always tangent adverse off V Y over Vieques. It's the same thing for the angle, for the acceleration is just the tangent adverse of a Y over X was nothing really new there. Okay, so let's look at the components now of the acceleration vector Now for the velocity components. We also had two equations. Remember velocities always displacement over time, so velocity and the X direction was either calculated. If you have Delta X and Delta T or if you had the magnitude and direction off that two dimensional vectors, those two different equations well, it's the same kind of idea for the acceleration vector. So it's two different equations. Remember that if the two dimensional acceleration is change in velocity over change in time, then the acceleration in the X direction is just the change in velocity in the X direction over time and the change in velocity and the Y is gonna be the change in velocity in the Y direction over change in time. So it's the same idea. There now we could also calculate this A these a X and A Y components by using the magnitude in the direction of the acceleration vector. So if we have a and theta than a X and Y are just a co sign data and a sign data. Alright, guys. So the equations are very similar against really just the letters. They're different, they work the same exact way. So let's just go ahead and jump right into an example. All right, so we have a toy car that's moving initially at 20 m per second and then later on it's moving at at some angle. So in this first part of the problem, we're gonna calculate the X and Y components off the cars acceleration. So for part, A were asked for a X and A Y. But before we get to that, I just want to draw a quick sketch of what's going on here. So we have this toy car that is initially moving, so this is my initial velocity At 20 m per second. It's purely on the X axis like this. So basically, this is my initial. This is my initial, and then my final is where the velocity is not moving just along the X axis. Now it's actually moving at some angle above, so it's basically moving at some angle like this. We know the final velocity here, 67 we know this angle here is 26.5 degrees. So there's a change in velocities. The magnitude and the direction both change. So there were some acceleration. So how do we calculate the components off this acceleration Vector over here. Let's just go through our equations. Remember, we have two equations. We could either use the change in the velocity and time, or we can use the magnitude and the direction of the of the acceleration vector. But if we take a look here, we actually don't have the acceleration, vector, the magnitude or the direction. The only thing we have is the magnitude and direction of the final velocity vector. They're not the same thing to don't get confused there. So let's see. I also I'm told some information about the time I know that's 10 seconds, and I have the initial and final velocities here, so I don't have any of these, Uh, you know, I don't have the magnitude of direction, but I might be able to figure out what the change in velocity in the X and Y directions are by using the initial and the final. Somebody used these equations over here. So my ex component is gonna be the change in velocity in the X direction over changing time. So it's basically just the final in the X minus. The initial in the X, divided by Delta T and then a Y is gonna be very similar. So it's gonna be changing velocity in the Y direction over changing time or velocity final minus medalist Initial in the UAE divided by Delta T If I can figure out basically with all these components are off these initial and final velocities and I can figure out the acceleration. So how did I do that? Let's take a look at my initial velocity vector. It's purely in the X direction, which means that technically, it's a one dimensional vector. But that also means I know the components. So I know that the X direction component is just 20 m per second. It's all in the basically in the X axis, and there's no components that lies in the Y axis. So that means that my V initial in the why is 0 m per second. It's not moving in the y axis, but the final velocity actually does point in both directions. So I'm gonna have to break it down into its X and Y components. So this is gonna be my v y. And this is gonna be my V X. And how do I get the V y and V X? Well, I actually have the magnitude and the direction off that two dimensional velocity vector, so I'm just gonna use these equations over here. Vico signed data and be signed data. So my V final in the X direction is gonna be my V final times, the cosine of data. So it's basically just gonna be 67 times the cosine of 26.5 and I get 60. So I know VFW final. The X is 60. My v y is gonna be yeah, in a very similar way 67 times the sign of 0.5 and I get 30. So I know my V final in the Y direction is 30. So now I actually have all the components that I need to solve the acceleration equation. I know that this is gonna be my V final. The X is 60. My the initial in the X is 20/10. So my acceleration component in the X direction is 4 m per second squared now, in a similar way, the V fight on the why is 30 my The initial in the Y is zero. Remember, its not moving along the X or y, uh, sorry along the Y direction. So 30 minus 0/10 is gonna be 3 m per second squared. So these are my acceleration components four and three. So we're done with party. So now moving into part B, we're gonna calculate the magnitude and the direction of the cars acceleration. So this is basically gonna be what's the magnitude of a and what is the direction of a So we know that the that the let's take a look at our equations, right? So we have two different ways of calculating the magnitude of a We can use the change in velocity over change in time. Remember, this is the change in velocity in two dimensions, or we could just use the Pythagorean theorem if we know the components. So if you take a look here, we actually just calculated the components a X and A y. So we're just gonna use the Pythagorean theorem. So are a is just gonna be the Pythagorean theorem of a squared in a Y squared. So this is just, uh this is gonna be the square root of four squared plus three squared, and we know that's equal to 5 m per second squared. So that is the magnitude of the acceleration. Now, if the direction we're just gonna use the tangent in verse off and then we're gonna use 3/4 and we have to put the absolute value signs over there and this is gonna be 37 degrees, 37 degrees and that's the direction. Alright, guys, that's it for this one. Let me know if you ask any questions.

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Problem

A football at rest is kicked by a football kicker. The ball is in contact with the kicker’s foot for 0.050s, during which it experiences an acceleration a = 340 m/s^{2}. The ball is launched at an angle of 40° above the ground (x-axis). Calculate the horizontal and vertical components of the launch velocity.

A

13 m/s horizontal; 10.9 m/s vertical

B

130 m/s horizontal; 109 m/s vertical

C

10.9 m/s horizontal; 13 m/s vertical

D

17 m/s horizontal; 17 m/s vertical

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