Intro to Motion in 2D: Position & Displacement: Videos & Practice Problems

Motion in two dimensions involves combining two one-dimensional motions, allowing movement at an angle. Position is represented as a vector, denoted as 𝛠, with coordinates (x, y). Displacement, the shortest path between two points, is calculated using the Pythagorean theorem: δr=√(δx^+δy^) and direction via θ=tan−1(|δy/δx|). Understanding these concepts is crucial for analyzing two-dimensional motion effectively.

Intro to Motion in 2D

Understanding motion in a two-dimensional plane, often referred to as motion at an angle, is essential for solving various physics problems. Unlike one-dimensional motion, which is confined to either the x-axis (horizontal) or the y-axis (vertical), two-dimensional motion allows for movement in both directions simultaneously. This concept can be visualized by considering a scenario where an object moves from point A to point C at an angle θ, rather than following a straight path along just one axis.

In two-dimensional motion, the movement can be decomposed into two separate one-dimensional motions. For instance, if you were to move from point A to point B and then from point B to point C, you would effectively be combining these two motions. This approach leads to the formation of triangles, which are fundamental in vector analysis.

To analyze motion in two dimensions, it is crucial to break down the movement into its x and y components. This involves applying the principles of average velocity and uniformly accelerated motion (UAM) equations alongside vector equations. By utilizing these combinations, you can effectively solve problems involving two-dimensional motion, treating them as a series of interconnected triangles. This method not only simplifies the analysis but also enhances your understanding of how different motion components interact.

Position and Displacement in 2D

In two-dimensional motion, position and displacement are represented as vectors, which can be visualized as arrows pointing from the origin to a specific point in a coordinate system. The position vector, denoted as \(\vec{r}\), is defined by its coordinates \((x, y)\), indicating where an object is located in the plane. For example, a position vector might be represented as \(\vec{r}_A\) at a distance of 3.6 units at an angle of 33.7 degrees from the origin.

To analyze these vectors, we can decompose them into their x and y components using trigonometric functions. The equations for the components are given by:

\(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).

For instance, if \(\vec{r}_A\) is 3.6 units at 33.7 degrees, the x-component would be calculated as:

\(x_A = 3.6 \cos(33.7^\circ) \approx 3\) and \(y_A = 3.6 \sin(33.7^\circ) \approx 2\).

Similarly, for another position vector \(\vec{r}_B\) with a magnitude of 8.49 units at 45 degrees, the components would be:

\(x_B = 8.49 \cos(45^\circ) \approx 6\) and \(y_B = 8.49 \sin(45^\circ) \approx 6\).

Displacement, on the other hand, is defined as the shortest path between two points, represented as \(\Delta \vec{r}\). It is calculated as the change in position, which can be expressed as:

\(\Delta x = x_B - x_A\) and \(\Delta y = y_B - y_A\).

Using the previous example, if \(x_A = 3\) and \(x_B = 6\), then:

\(\Delta x = 6 - 3 = 3\) and \(\Delta y = 6 - 2 = 4\).

The magnitude of the displacement can be found using the Pythagorean theorem:

\(|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}\).

Substituting the values gives:

\(|\Delta \vec{r}| = \sqrt{3^2 + 4^2} = 5\) meters.

The direction of the displacement is determined using the inverse tangent function:

\(\theta = \tan^{-1}\left(\frac{|\Delta y|}{|\Delta x|}\right)\).

In this case, it results in:

\(\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53^\circ\).

In summary, understanding two-dimensional vectors involves breaking them down into their components and applying trigonometric relationships, while displacement is calculated as the change in position, utilizing both the Pythagorean theorem and inverse tangent for magnitude and direction, respectively.

At point A, a hiker is 10m east from the origin. After 35s, the hiker arrives at point B 40m at 60° north of east from the origin. Calculate the magnitude and direction of the hiker's displacement.

• A. 50m; 60° north of east
• B. 36m; 73.9° north of east
• C. 36m; 60° north of east
• D. 36m; 16.1°north of east

Answer: (B)

Final Position Vector

In this problem, we are tasked with determining the final position vector after a series of movements starting from an initial position. The initial position is given as a vector with a magnitude of 6.2 units at an angle of 25 degrees below the x-axis. This means that the vector points into the fourth quadrant of the coordinate system. To visualize this, we can represent the initial position vector as R_A.

Next, we travel a distance of 9.9 units at an angle of 78 degrees above the positive x-axis, which we will label as R_AB. Finally, we move 2 meters in the negative x direction, represented as R_BC. Our goal is to find the magnitude and direction of the final position vector R_C.

To solve this, we first break down each vector into its x and y components. The components of a vector can be calculated using the following equations:

x = r * cos(θ)

y = r * sin(θ)

For the initial position vector R_A:

x_A = 6.2 * cos(25°) = 5.62

y_A = 6.2 * sin(25°) = -2.62 (negative because it is below the x-axis)

For the displacement vector R_AB:

x_AB = 9.9 * cos(78°) = 2.06

y_AB = 9.9 * sin(78°) = 9.68

For the displacement vector R_BC (which only affects the x-component):

x_BC = -2

y_BC = 0

Now, we can compile these components into a table to sum them up:

Vector	x Component	y Component
RA	5.62	-2.62
RAB	2.06	9.68
RBC	-2	0

Adding the x-components:

x_C = 5.62 + 2.06 - 2 = 5.68

Adding the y-components:

y_C = -2.62 + 9.68 + 0 = 7.06

Now that we have the components of the final position vector R_C, we can calculate its magnitude using the Pythagorean theorem:

|R_C| = √(x_C² + y_C²) = √(5.68² + 7.06²) = 9.06

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

θ_C = tan^-1(y_C / x_C) = tan^-1(7.06 / 5.68) = 51.2°

Thus, the final position vector R_C has a magnitude of 9.06 units and is directed at an angle of 51.2 degrees above the positive x-axis.

Here’s what students ask on this topic:

What is the difference between position and displacement in 2D motion?

In 2D motion, position is a vector that specifies the location of a point in a coordinate system, represented as $r_{}$ with coordinates (x, y). It points from the origin to the point of interest. Displacement, on the other hand, is the shortest path between two points and is represented as ${\mathrm{\delta r}}_{}$. It is calculated using the Pythagorean theorem: ${\mathrm{\delta r}}_{}=\sqrt{{{\mathrm{\delta x}}_{}}^{}+{{\mathrm{\delta y}}_{}}^{}}$ and its direction is given by $\theta ={\tan }^{-1}(|{\mathrm{\delta y}}_{}/{\mathrm{\delta x}}_{}|)$. Position is about where you are, while displacement is about how far and in what direction you have moved from one point to another.

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

To calculate the x and y components of a position vector in 2D motion, you use trigonometric functions. If the position vector $r_{}$ has a magnitude $r$ and makes an angle $\theta$ with the x-axis, the components are given by: $x_{}=r\times \cos \theta$ and $y_{}=r\times \sin \theta$. For example, if $r_{}=\; 3.6$ and $\theta =\; 33.7°$, then $x_{}=\; 3.6\times \cos 33.7°\; \approx \; 3$ and $y_{}=\; 3.6\times \sin 33.7°\; \approx \; 2$.

What is the significance of breaking down 2D motion into x and y components?

Breaking down 2D motion into x and y components simplifies the analysis of motion. It allows us to use one-dimensional motion equations separately for each axis. This approach leverages vector decomposition, where a vector is split into perpendicular components, making it easier to apply trigonometric and algebraic methods. By analyzing the x and y components independently, we can solve complex problems involving angles and trajectories using simpler, well-understood principles of one-dimensional motion.

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

To determine the magnitude and direction of displacement in 2D motion, you first find the changes in the x and y coordinates, ${\mathrm{\delta x}}_{}$ and ${\mathrm{\delta y}}_{}$. The magnitude of displacement ${\mathrm{\delta r}}_{}$ is calculated using the Pythagorean theorem: ${\mathrm{\delta r}}_{}=\sqrt{{{\mathrm{\delta x}}_{}}^{}+{{\mathrm{\delta y}}_{}}^{}}$. The direction $\theta$ is found using the inverse tangent function: $\theta ={\tan }^{-1}(|{\mathrm{\delta y}}_{}/{\mathrm{\delta x}}_{}|)$. For example, if ${\mathrm{\delta x}}_{}=\; 3$ and ${\mathrm{\delta y}}_{}=\; 4$, then ${\mathrm{\delta r}}_{}=\sqrt{3{}^{}+\; 4{}^{}}=\; 5$ and $\theta ={\tan }^{-1}(4/3)\; \approx \; 53°$.

Why is understanding vector decomposition important in 2D motion?

Understanding vector decomposition is crucial in 2D motion because it allows us to break down complex movements into simpler, manageable components. By decomposing vectors into their x and y components, we can apply basic trigonometric and algebraic methods to solve problems. This approach simplifies the analysis of motion, making it easier to calculate positions, velocities, and accelerations in each direction independently. It also helps in visualizing and understanding the overall motion by considering the contributions of each component separately.