Velocity in 2D - Online Tutor, Practice Problems & Exam Prep

In two-dimensional motion, speed is a scalar quantity defined as distance over time, while velocity is a vector quantity represented as displacement over time. The equations for calculating average speed and velocity involve determining the total distance traveled and the shortest path (displacement). For example, if walking 40 meters right and 30 meters up in 10 seconds, average speed is calculated as $\frac{d}{\mathrm{\Delta t;}}$ resulting in 7 m/s, while velocity magnitude is $\frac{\mathrm{\Delta r}}{\mathrm{\Delta t;}}$ yielding 5 m/s at an angle of 37 degrees.

Average Speed and Velocity in 2D

Hey, guys. So now that we've seen how displacement works in 2 dimensions, in this video, I want to cover how speed and velocity work in 2D. What we're going to use is a lot of the same equations that we use for 1-dimensional motion. But there are a couple of differences, so let's talk about those and then do a quick example. 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 going to be at some angle rather than just on a flat line. But the idea is the same. So the difference between speed and velocity is that speed is a magnitude only and it's a distance over time. So d/Δt and distance is a scalar, so therefore speed is a scalar. 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 Δx/Δt. But now we're working in 2D. So we just use Δr. That's really all there is to it. So the magnitude is going to be Δr/Δt. So now we're going to have displacement vectors that point off at some angle like this. This is Δr and this is the angle theta. So what we also have to do now is specify the direction and the direction is given by an angle θ_v. And so here's the deal: If your velocity is given as the displacement over time, Δr/Δt, then whatever direction your Δr points off in, that's going to be the same exact direction as your velocity vector. So basically, your velocity always points in the same direction as Δr. And so whatever θ_v is, it's the same thing as theta for Δr. They share the same angle. That's really all there is to it. Let's go ahead and do an example.

So we're going to walk 40 in the x-axis and then 30 in the y-axis and the trip takes 10 seconds. So we know this is 40 and this is 30 and we know this is our delta T. So we're going to calculate the average speed in the first part and then the magnitude and direction of the velocity. For the first part, going to calculate the speed. That's s. And remember that is equal to, distance over time. So it's d/Δt. We know the Δt is 10 seconds. Now we just need to figure out the distance. So remember, I'm walking 40 in the x and 30 in the y. These are both the lengths that I'm traveling. And so that means that the distance is just the sum of all the lengths that I'm traveling. So that's 40 plus 30 and that's 70. So that means that my speed is going to be 70 over 10, and that is 7 meters per second on average. So that's my average speed. So let's move on now to the velocity. I want to figure out the magnitude of the velocity and I want to figure out the angle. So the magnitude is going to be Δr/Δt. That's the 2-dimensional displacement. So again, we know that Δt is 10 seconds, but now we just need to go and figure out my 2-dimensional displacement vector. So remember, I'm walking 40 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 Δr. And so how do we figure out the magnitude of this displacement? Well, this is just a triangle and we have the legs. This is 30 and 40. So this is a 3, 4, 5 triangle and that means Δr is 50 meters. Even though I literally want 70, my displacement is only 50 meters from start to finish. So that means that the magnitude of my velocity is going to be 50 meters over 10 seconds. So this is 5 meters per second. Now for the angle. Well, basically, the angle over here is going to be this guy, this angle over here. And so remember that the angle for your velocity vector is going to be the same thing as the angle for your displacement. So I've got the displacement vector and I know that the velocity is going to point off in this direction. So this is my V, and I know that they're going to share the same angle theta. So what I can do is I can just set up my tangent inverse equation because that's the theta, and I'm just going to use the legs of the triangle. So I'm going to use basically delta y over delta x. And so this is going to be tangent inverse of 30 over 40, and that's going to be 37 degrees. So that is the magnitude and the direction. Alright, guys. That's all there is to it. Let's go ahead and get some more practice.

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

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

Calculating Velocity Components

Hey, guys. So in the last couple of videos, I introduced you to the velocity vector in 2 dimensions. So if the velocity is ever 2D, then just like any other vector, it has x and y components. In this video, I'm going to show you how to calculate those velocity components. There are really 2 sets of equations that we're going to use to jump back and forth between the velocity and its components. It really just comes down to what problems we'll give you. So basically, there are 2 different kinds of scenarios. You're either going to be going back and forth between the components and displacement in time, or you're going to go back and forth between the components and the magnitude and direction of the velocity vector. What we're going to see is that once we fill out this table, once 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 going to walk 40 meters to the right, 30 meters up in 10 seconds, and we're going to calculate the magnitude and the x and y components of the velocity. So here, we're going to go back and forth between the components and time. So let's go ahead and do that. Right? So we're going to go 40 meters to the right and then 30 meters up, and we're going to calculate the magnitude of the velocity vector. Well, we've seen that equation before. Remember, it's just Δr/Δt (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 2-dimensional displacement. Well, I'm going 40 to the right and 30 up, so it means my 2-dimensional displacement, Δr, is going to be the hypotenuse of this triangle. So we're just going to use the Pythagorean theorem. We're just going to use some triangle equations. And really, this is a 3, 4, 5 triangle. So if this is 30 and 40, the legs of the triangle, then that means the hypotenuse is 50. So this is 50 meters. That's our 2-dimensional displacement. And so our velocity is just going to be 5 meters per second. That's the magnitude. Now remember that this velocity vector also points in the same direction as Δr. So that means the velocity vector points in this direction. So we know that v is 5. We're not quite done yet because remember, we have to calculate the velocity's magnitude and its x and y components. So we figured out the first part. Now we're just going to figure out the x and y components. So this velocity vector over here is 2-dimensional. It points at some angle like this. So, we can break it down into a triangle just as we would any other vector. So this is my v_x component and this is my v_y component over here. So we're going to figure out v_x and v_y. So what are the equations we're going to use for that? Well, remember that velocity is always displacement over time. In two dimensions, v was just Δr/Δt. But now we're looking for v_x, the component of the velocity in the x direction. But it's still going to be displacement over time. It's just going to be the displacement in the x direction over time. So it's Δx/Δt. And in a similar way, v_y is going to be the displacement in the y direction over change in time. So it's always displacement over time. That's always going to be what velocity is. Alright. So that means that we're going to use Δx/Δt. So what's our Δx? Well, that's 40, so we're going to use 40 over 10 seconds and this is going to be 4 meters per second. So that's the components in the x direction. You can think of this as how much of this 5 that lies in 2 dimensions lies basically just along the x axis and that's 4. So our v_y components are going to be Δy/Δt. So this is going to be 30 over 10 and this is going to be 3 meters 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 3-4-5 triangle with the displacements, we also ended up with a 3-4-5 triangle with the velocity. And it's because really all of 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 going to be walking at 5 meters per second at some angle 37 degrees above the x-axis, and we're going to calculate the x and y components of the velocity. So in this particular problem, we're given a velocity in 2 dimensions. We go in the magnitude and the angle here, and we're going to calculate the x and y components. So, basically, this is just going to break down into a triangle just like any other 2-dimensional vector. So now this is going to 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 going to be triangle equations. These are just going to be vector equations. So in general, what happens is the magnitude of this velocity is just going to be the hypotenuse of the triangle. So this is just going to be your Pythagorean Theorem, v_x² + v_y². Your angle θ is going to be the tangent inverse of the absolute value of v_y over v_x. And if we want to calculate the x and y components, then we're just going to use v cosine θ and v sine θ. 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. Alright. So that means that your velocity component v_x is just going to be v times cosine θ, which is just going to be in this example, 5 times the cosine of 37 and this is going to be 4 meters per second. So this is the components of the velocity in the x direction. And then my d_y is going to be very similar. It's going to be 5 times the cosine of 37 and that's going to be 3 meters per second. So let me write that a little bit bigger. So this is going to be 3 meters per second. Alright? So this is the components of my x and y velocities. So notice here, the last thing I want to mention, is that we've ended up with the 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 used 2 different setups, 2 different equations, but these really just describe the same exact problem here. Alright? So let me know if you guys have any questions. That's it for this video.

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

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?

• A. (11.8m, 13.6m)
• B. (25.8m, 35.6m)
• C. (41.8m, 43.6m)
• D. (33.8m, 27.6m)

Here’s what students ask on this topic:

What is the difference between speed and velocity in 2D motion?

In 2D motion, speed is a scalar quantity that measures how fast an object is moving, calculated as the total distance traveled divided by the time taken. The formula for speed is $\frac{d}{\mathrm{\Delta t}}$. On the other hand, velocity is a vector quantity that includes both magnitude and direction. It is calculated as the displacement (the shortest path between the starting and ending points) divided by the time taken. The formula for velocity is $\frac{\mathrm{\Delta r}}{\mathrm{\Delta t}}$. For example, if you walk 40 meters to the right and 30 meters up in 10 seconds, your average speed would be 7 m/s, while your velocity would be 5 m/s at an angle of 37 degrees.

How do you calculate the magnitude and direction of velocity in 2D motion?

To calculate the magnitude and direction of velocity in 2D motion, you first need to determine the displacement vector $\mathrm{\Delta r}$. For example, if you walk 40 meters to the right and 30 meters up, the displacement is the hypotenuse of the right triangle formed, which can be calculated using the Pythagorean theorem: $\sqrt{{40}^2+{30}^2}=\; 50$ meters. The magnitude of velocity is then $\frac{50}{10}=\; 5$ m/s. The direction is given by the angle $\theta$, which can be found using the inverse tangent function: $tan^-1\frac{30}{40}=\; 37$ degrees.

How do you find the x and y components of a velocity vector in 2D motion?

To find the x and y components of a velocity vector in 2D motion, you can use trigonometric functions. If you know the magnitude of the velocity $v$ and the angle $\theta$ it makes with the x-axis, the x-component $\mathrm{v_x}$ is given by $\mathrm{v_x\; =\; v\; cos(\theta )}$, and the y-component $\mathrm{v_y}$ is given by $\mathrm{v_y\; =\; v\; sin(\theta )}$. For example, if the velocity is 5 m/s at an angle of 37 degrees, the x-component is $\mathrm{5\; cos(37°)\; =\; 4}$ m/s, and the y-component is $\mathrm{5\; sin(37°)\; =\; 3}$ m/s.

What equations are used to convert between velocity components and magnitude/direction in 2D motion?

To convert between velocity components and magnitude/direction in 2D motion, you use the following equations: To find the magnitude $v$ from components $\mathrm{v_x}$ and $\mathrm{v_y}$, use the Pythagorean theorem: $\sqrt{{\mathrm{v_x}}^2+{\mathrm{v_y}}^2}$. To find the direction $\theta$, use the inverse tangent function: $tan^-1\frac{\mathrm{v_y}}{\mathrm{v_x}}$. To find the components from magnitude and direction, use $\mathrm{v_x\; =\; v\; cos(\theta )}$ and $\mathrm{v_y\; =\; v\; sin(\theta )}$. These equations allow you to switch between different representations of velocity in 2D motion.

How do you calculate average speed and velocity in a 2D motion problem?

To calculate average speed in a 2D motion problem, you need the total distance traveled and the total time taken. The formula is $\frac{d}{\mathrm{\Delta t}}$. For example, if you walk 40 meters to the right and 30 meters up in 10 seconds, the total distance is 70 meters, so the average speed is $\frac{70}{10}=\; 7$ m/s. To calculate average velocity, you need the displacement (the shortest path between the starting and ending points) and the total time taken. The formula is $\frac{\mathrm{\Delta r}}{\mathrm{\Delta t}}$. Using the same example, the displacement is 50 meters (calculated using the Pythagorean theorem), so the average velocity is $\frac{50}{10}=\; 5$ m/s at an angle of 37 degrees.