Velocity in 2D - Video Tutorials & Practice Problems

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1

concept

Average Speed and Velocity in 2D

Video duration:

4m

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Hey, guys. So now that we've seen how displacement works in two dimensions in this video, I want to cover how speed and velocity work in two D what we're gonna uses a lot of the same equations that we use for one dimensional motion. But there's a couple of differences. So let's talk about those and to do a quick example. So just remember that speed and velocity just measure how fast something is moving between two points. The only thing that's new here is that these two points are gonna be at some angle rather than just on flat line. But the idea is the same. So the difference between speed and velocity is that speed is a magnitude on Lee, and it's a distance over time. So Dior, Delta T and Distance is a scaler, so therefore, speeds the scaler. On the other hand, velocity is a magnitude and a direction. It's not distance over time. It's displacement over time. Displacement is a vector, and so therefore, velocity is also a vector. So the only thing that's new here, guys, is that in one dimension we use the one dimensional displacement Delta X over Delta T. But now we're working in two D. So we just use Delta. Are that's really all there is to it. So the magnitude is gonna be Delta are over Delta T. So now we're gonna have displacement vectors that point off its some angle like this. This is Delta are and this is the angle theta. So we also have to do now is specified the direction and the directions given by an angle fate of V And so here's the deal. If your velocity is given as the displacement overtime Delta, our delta T then what? Our direction your delta are points often that's gonna be the same exact direction as your velocity vector. So basically your velocity always points in the same direction as Delta are. And so whatever Theta V is, it's the same thing as data for Delta are they share the same angle. That's really all there is to it. Let's go ahead and do an example. So we're gonna walk 40 in the X axis and then 13 y, and the trip takes 10 seconds. So we know this is 40 and we know this is 30 and we know this is our delta t so we're gonna calculate the average speed in the first part and then the magnitude and direction of the velocity. So for the first part, we're gonna calculate the speed that's s and remember that is equal to, uh, distance over time. So it's d over Delta T. We know the Delta T is 10 seconds now. We just need to figure out the distance. So remember, I'm walking 40 in the X and 30 and the why these air both links that I'm traveling. And so that means that the distance is just the sum of all the links that I'm traveling. So that's 40 plus 30 and that's 70. So that means that my speed, this is gonna be 70/10 and that is 7 m per second on average. So that's my average speed. So let's move on down to the velocity. Want to figure out the magnitude of the velocity and I want to figure out the angle so the magnitude is gonna be Delta are over Delta T. That's the two dimensional displacement. So again, we know the Delta T is 10 seconds, but now we just need to go and figure out my two dimensional displacement vector. So remember, I'm walking 40 and 30 so I'm walking literally 70 in terms of distance. But my displacement is the shortest path from start to finish. So that's this angle or this vector over here. This is Delta are. And so how do we figure out the magnitude of this displacement? Well, this is just a triangle. We have the legs. This is 30 and 40. So this is a 345 triangle and that means Delta R is 50 m. So even though I literally walked 70 my displacement is only 50 m from start to finish. So that means that my the magnitude of my velocity is gonna be 50 m over seconds. So this is five meters per second now for the angle. Well, basically, the angle over here is gonna be this guy, this angle over here on dso remember that the angle for your velocity vector is gonna be the same thing as the angle for your displacement. So I've got the displacement vector, and I know that the velocity is gonna point off in this direction. So this is my V. And I know that they're going to share the same angle, Fada. So what I can do is I can just set up my tangent inverse equation, because that's the data. And I'm just gonna use the legs of the triangle. So I'm gonna use basically Delta y over Delta X. And so this is gonna be tangent inverse of over 40 and that's gonna be 37 degrees. So that is the magnitude and direction. Alright, guys, that's all there is to it. Let's go ahead and get some more practice.

2

Problem

Problem

While following a treasure map, you start at an old oak tree. You first walk 85 m at 30.0° west of north, then walk 92 m at 67.0° north of east. You reach the treasure 2 minutes later. Calculate the magnitude of your average velocity for the entire trip.

A

1.11 m/s

B

1.48 m/s

C

1.40 m/s

D

1.32 m/s

3

Problem

Problem

While following a treasure map, you start at an old oak tree. You first walk 85 m at 30.0° west of north, then walk 92 m at 67.0° north of east. You reach the treasure 2 minutes later. Calculate your average speed for the entire trip.

A

1.5 m/s

B

177 m/s

C

88.5 m/s

4

concept

Calculating Velocity Components

Video duration:

5m

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Hey, guys. So in the last couple videos, I introduce you the velocity vector into dimensions. So if the velocity is ever to d then just like any other vector, it has X and y components. So in this video, I'm gonna show you how to calculate those velocity components. And there's really two sets of equations that we're gonna use to jump back and forth between the velocity and its components and really just comes down to what problems will give you. So basically, there's two different kinds of scenarios. You're either gonna be going back and forth between the components and displacement and time, or you're gonna go back and forth between the components and the magnitude and direction of the velocity vector. And really, what we're gonna see is that once we fill out this table, what's thes? We fill out these equations. These are all just vector and motion equations that we've seen before. Let's go ahead and check it out here. So let's take a look at the first example. We're gonna walk 40 m to the right, 30 m up in 10 seconds, and we're gonna calculate the magnitude and the X and Y components of the velocity. So here we're gonna go back and forth between the components and displacement and time. So let's go ahead and do that right. So we're gonna go 40 m to the right and then 40 and then 30 m up and we're gonna calculate the magnitude of the velocity vector. What? We've seen that equation before, remember, That's just Delta are over Delta T displacement. Over time. We've seen that equation before, so I know that my time is 10 seconds. And now all I have to do is just calculate my two dimensional displacement. Well, im going 40 to the right and 30 up. So that means my two dimensional displacement Delta are is gonna be the high pot news of this triangle. So we're just gonna use Pythagorean theorem. We're just gonna use, um, triangle equations. And really, this is a 345 triangle. So if this is 30 and 40 the legs of the triangle, then that means the high pot news is 50. So this is 50 m. That's our two dimensional displacement. And so our velocity is just gonna be 5 m per second. That's the magnitude. Now, remember that this velocity vector also points in the same direction as Delta are. So that means the velocity vector points in this direction. So we know that V is five. But we're not quite done yet because remember, we have to calculate the velocities magnitude, and it's X and Y components. So we figure out the first part now we're just gonna figure out the X and Y components, So this velocity vector over here is two dimensional. It pointed some angle like this so we can break it down into a triangle just like we would any other vector. So this is my V X component, and this is my V y component over here. So we're gonna figure out VX and VY y. So what do the equations we're gonna use for that? Well, remember that velocity is always displacement Overtime in two dimensions. V was just Delta are over Delta T. But now we're looking for Vieques, the component of the velocity in the X direction, But it's still gonna be displacement over time. It's just gonna be the displacement in the X direction over time. So it's Delta X over Delta T and in a similar way V Y is gonna be the displacement in the Y direction over change in time. So it's always displacement over time. That's always gonna be what velocity is. All right, So that means we're gonna use Delta X over Delta T. So what's our Delta X? Well, that's 40. So we're gonna use 40/10 seconds and this is gonna be 4 m per second. So that's the component in the X direction. You could think of this as how much of this five that lies in two dimensions lies basically just along the X axis. And it's four. So our B Y components is gonna be Delta y over Delta T. So there's gonna be 30/10 and this is gonna be 3 m per second. So this is basically how much of this vector lies in this direction and sort of like the vertical axis like this. Alright, So notice also how, just like we had a 345 triangle with the displacements. We also ended up with a 345 triangle with the velocity. And it's because, really, all these numbers are getting divided by the time which is 10 seconds. Alright, so Let's move on. Now let's take a look at the second example. So in this example, we're gonna be walking at 5 m per second at some angle 37 degrees above the X axis on, we're gonna calculate the X and Y components of the velocity. So in this particular problem, were given a velocity in two dimensions were given the magnitude and the angle here, and we're gonna calculate the X and Y components. So basically, this is just gonna break down into a triangle just like any other two dimensional vector. So now this is gonna be my V x and my V Y components here. So basically, the relationship between all these variables the magnitude, the direction and your components are all just gonna be triangle equations. These they're just gonna be vector equations. So in general, what happens is the magnitude of this velocity is just gonna be the hypothesis of the triangle. So this is just gonna be your Pythagorean theorem. V X squared plus v y squared your angle theta is gonna be the tangent inverse of the absolute value of ey over Vieques. And if we want to calculate the X and y components, then we're just gonna use V cosign data and V sign data So we can see here that all of these equations are either just motion equations that we know your displacement over time or they're just vector equations that we already know. All right, So that means that your velocity component V X is just gonna be v times the cosine data, which is just gonna be in this example five times the coastline of 37. And this is gonna be 4 m per second. So this is the components of the velocity in the X direction, and then my d y is gonna be very similar. It's gonna be five times the sign of 37 and that's gonna be 3 m per second. So let me write that a little bit bigger, so there's gonna be three meters per second. All right, so this is the components of my X and Y velocities. So notice here last week I mentioned is that we've ended up with same exact numbers that we did on this right example over here, as we did on the left, we ended up with the same exact numbers. So we use two different setups, two different equations. But he's really just described the same exact problem here. All right, so let me know if you guys have any questions, that's it for this one.

5

Problem

Problem

A coastal breeze pushes your sailboat at constant velocity for 8 min. After checking your instruments, you determine you've been pushed 650 m west and 800 m south. What was the magnitude & direction of your average velocity?

A

2.15 m/s; 39.1° south of west

B

128.9 m/s; 50.9° south of west

C

2.15 m/s; 50.9° south of west

6

Problem

Problem

A ball moves on a tabletop. The ball has initial x & y coordinates (1.8m, 3.6m). The ball moves 10m/s at 53.1° above the x-axis for 4s. What are the x & y coordinates of the ball's final position?