Velocity in 2D: Videos & Practice Problems

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

Understanding speed and velocity in two dimensions builds on the concepts established in one-dimensional motion, but with a focus on the direction of movement. Speed is defined as the distance traveled over time, represented by the formula:

$$ s = \frac{d}{\Delta t} $$

Here, distance is a scalar quantity, meaning it only has magnitude. In contrast, velocity incorporates both magnitude and direction, calculated using displacement rather than distance:

$$ v = \frac{\Delta r}{\Delta t} $$

Displacement, a vector quantity, accounts for the shortest path between two points and includes directional information. In two dimensions, displacement is represented as a vector that can be expressed in terms of its components along the x and y axes, denoted as Δx and Δy. The overall displacement vector can be represented as Δr, which points at an angle θ from the origin.

When calculating average speed and velocity, it is essential to differentiate between the total distance traveled and the net displacement. For example, if an object moves 40 meters along the x-axis and 30 meters along the y-axis over a time interval of 10 seconds, the total distance traveled is:

$$ d = 40 + 30 = 70 \text{ meters} $$

Thus, the average speed is:

$$ s = \frac{70 \text{ m}}{10 \text{ s}} = 7 \text{ m/s} $$

To find the magnitude of the velocity, we first determine the displacement using the Pythagorean theorem, as the movement forms a right triangle:

$$ \Delta r = \sqrt{(40^2 + 30^2)} = \sqrt{(1600 + 900)} = \sqrt{2500} = 50 \text{ meters} $$

Consequently, the magnitude of the velocity is:

$$ v = \frac{50 \text{ m}}{10 \text{ s}} = 5 \text{ m/s} $$

To find the direction of the velocity vector, we can use the tangent function, which relates the opposite side (Δy) to the adjacent side (Δx) of the triangle:

$$ \theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right) = \tan^{-1}\left(\frac{30}{40}\right) = 37^\circ $$

This angle indicates the direction of the velocity vector, which aligns with the direction of the displacement vector. Thus, in two-dimensional motion, both speed and velocity are crucial for understanding the movement of objects, with velocity providing essential directional context.

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

Answer: (D)

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

Answer: (A)

Calculating Velocity Components

In two-dimensional motion, the velocity vector can be broken down into its x and y components, which are essential for analyzing movement in a plane. To calculate these components, we can use two primary approaches depending on the information provided: either through displacement and time or through magnitude and direction.

When given displacement in both the x and y directions, we can determine the overall displacement using the Pythagorean theorem. For example, if an object moves 40 meters to the right and 30 meters up, the total displacement (\(\Delta r\)) can be calculated as:

\(\Delta r = \sqrt{(40^2 + 30^2)} = 50 \text{ meters}\)

To find the magnitude of the velocity (\(v\)), we divide the total displacement by the time taken. If the time is 10 seconds, the velocity is:

\(v = \frac{\Delta r}{\Delta t} = \frac{50 \text{ m}}{10 \text{ s}} = 5 \text{ m/s}\)

Next, to find the x and y components of the velocity (\(v_x\) and \(v_y\)), we use the formulas:

\(v_x = \frac{\Delta x}{\Delta t}\) and \(v_y = \frac{\Delta y}{\Delta t}\)

Substituting the values, we find:

\(v_x = \frac{40 \text{ m}}{10 \text{ s}} = 4 \text{ m/s}\)

\(v_y = \frac{30 \text{ m}}{10 \text{ s}} = 3 \text{ m/s}\)

This results in a 3-4-5 triangle for both displacement and velocity components, illustrating the relationship between the two-dimensional motion.

In another scenario, if we know the velocity's magnitude and angle, we can also find the components using trigonometric functions. For instance, if an object moves at 5 m/s at an angle of 37 degrees above the x-axis, the components can be calculated as:

\(v_x = v \cdot \cos(\theta) = 5 \cdot \cos(37^\circ) \approx 4 \text{ m/s}\)

\(v_y = v \cdot \sin(\theta) = 5 \cdot \sin(37^\circ) \approx 3 \text{ m/s}\)

These calculations yield the same component values as the previous example, demonstrating the consistency of vector decomposition methods.

Understanding how to transition between the magnitude and direction of a velocity vector and its components is crucial for solving problems in physics, particularly in kinematics. The relationships between these variables are foundational for analyzing motion in two dimensions.

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

Answer: (C)

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)

Answer: (B)

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.