Kinematics in 2D: Videos & Practice Problems

In two-dimensional motion with constant acceleration, problems can be approached similarly to one-dimensional motion by decomposing vectors into x and y components. For example, to find the displacement of a hockey puck affected by wind, one calculates the x and y displacements using the equations:

Δx = v_0x × t + 0.5 × a_x × t^2 and Δy = v_0y × t + 0.5 × a_y × t^2 The magnitude and direction of the resultant displacement are then determined using the Pythagorean theorem and trigonometric functions.

Kinematics in 2D

In solving kinematics problems involving two-dimensional motion, the approach mirrors that of one-dimensional motion, with the added complexity of vector decomposition. The key to tackling these problems is to break down the motion into its x and y components, allowing for the application of familiar equations of motion.

Consider a scenario where a hockey puck slides eastward at an initial velocity of 8 m/s, while being subjected to an acceleration of 3 m/s² at an angle of 37 degrees northeast. To determine the puck's displacement after 5 seconds, we first draw a diagram to visualize the vectors involved. The initial velocity vector is entirely in the x-direction, while the acceleration vector must be decomposed into its x and y components using trigonometric functions:

For the x-component of acceleration:

$$ a_x = a \cdot \cos(\theta) = 3 \cdot \cos(37^\circ) \approx 2.4 \, \text{m/s}^2 $$

For the y-component of acceleration:

$$ a_y = a \cdot \sin(\theta) = 3 \cdot \sin(37^\circ) \approx 1.8 \, \text{m/s}^2 $$

Next, we identify the known variables for both the x and y directions. In the x-direction, we have:

• Initial velocity, \( v_{0x} = 8 \, \text{m/s} \)
• Acceleration, \( a_x = 2.4 \, \text{m/s}^2 \)
• Time, \( t = 5 \, \text{s} \)

For the y-direction, the initial velocity is zero, and we have the acceleration and time. We can now apply the kinematic equation for displacement:

For the x-direction:

$$ \Delta x = v_{0x} t + \frac{1}{2} a_x t^2 $$

Substituting the known values:

$$ \Delta x = 8 \cdot 5 + \frac{1}{2} \cdot 2.4 \cdot (5^2) = 40 + 30 = 70 \, \text{m} $$

For the y-direction:

$$ \Delta y = v_{0y} t + \frac{1}{2} a_y t^2 $$

Since \( v_{0y} = 0 \):

$$ \Delta y = 0 + \frac{1}{2} \cdot 1.8 \cdot (5^2) = 0 + 22.5 = 22.5 \, \text{m} $$

With both displacements calculated, we can now find the magnitude of the resultant displacement vector using the Pythagorean theorem:

$$ |\Delta r| = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(70)^2 + (22.5)^2} \approx 73.5 \, \text{m} $$

To find the direction of the displacement, we use the inverse tangent function:

$$ \theta = \tan^{-1}\left(\frac{\Delta y}{\Delta x}\right) = \tan^{-1}\left(\frac{22.5}{70}\right) \approx 17.8^\circ $$

This angle indicates the direction of the displacement relative to the east, specifically north of east. Thus, the final results are a displacement magnitude of approximately 73.5 meters and a direction of 17.8 degrees north of east.

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?

• A. 2.8 m/s²; along –x axis
• B. 1.8 m/s²; 51.8° below –x axis
• C. 2.8 m/s²; above –x axis
• D. 1.3 m/s²; 24° above –x axis

Answer: (B)

Here’s what students ask on this topic:

What are the key steps to solve kinematics problems in 2D with constant acceleration?

To solve kinematics problems in 2D with constant acceleration, follow these steps: 1. Draw a diagram to visualize the problem. 2. Decompose any 2D vectors into x and y components. 3. List the five kinematic variables for both x and y directions: Δx, v_0x, v_fx, a_x, t and Δy, v_0y, v_fy, a_y, t. 4. Identify known and target variables. 5. Use kinematic equations to solve for the unknowns. For displacement, use Δx = v_0xt + 0.5a_xt² and Δy = v_0yt + 0.5a_yt². 6. Calculate the magnitude and direction of the resultant displacement using the Pythagorean theorem and trigonometric functions.

How do you decompose a 2D vector into x and y components?

To decompose a 2D vector into x and y components, use trigonometric functions. If the vector has a magnitude A and makes an angle θ with the x-axis, the x-component (A_x) is given by A_x = A cos(θ), and the y-component (A_y) is given by A_y = A sin(θ). This allows you to break down the vector into its horizontal and vertical components, which can be analyzed separately in kinematics problems.

What equations are used to find displacement in 2D kinematics?

In 2D kinematics, the displacement in the x and y directions can be found using the following equations: Δx = v_0xt + 0.5a_xt² and Δy = v_0yt + 0.5a_yt². These equations account for the initial velocity, acceleration, and time in each direction. The resultant displacement magnitude can be found using the Pythagorean theorem: Δr = √(Δx² + Δy²), and the direction can be determined using the inverse tangent function: θ = tan^-1(Δy/Δx).

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

To calculate the magnitude of the resultant displacement in 2D motion, use the Pythagorean theorem: Δr = √(Δx² + Δy²). For the direction, use the inverse tangent function: θ = tan^-1(Δy/Δx). Here, Δx and Δy are the displacements in the x and y directions, respectively. This will give you the angle θ relative to the x-axis, indicating the direction of the resultant displacement.

What is the importance of decomposing vectors in 2D kinematics problems?

Decomposing vectors in 2D kinematics problems is crucial because it allows you to analyze the motion separately in the x and y directions. By breaking down a 2D vector into its horizontal and vertical components, you can apply the kinematic equations independently to each direction. This simplifies the problem-solving process and ensures accurate calculations of displacement, velocity, and acceleration in two dimensions.