A Particle Undergoes Three Successive Displacements in a Plane

In physics, the motion of particles in a plane often involves a combination of different displacements. These displacements can be described using vectors, which have both magnitude and direction. Understanding how to break down these displacements and analyze their resultant vector is a crucial skill for solving problems in kinematics and dynamics.

In this article, we will delve into a specific problem: “A particle undergoes three successive displacements in a plane as follows: 4.0 m, 300 south of west; 5.0 m east; and 6.0 m, 600 north of east.” This problem requires us to break down each displacement into its vector components, calculate the resultant displacement, and determine its magnitude and direction. By thoroughly analyzing this problem, we aim to provide a deep understanding of vector operations in two dimensions.

Understanding the Problem Statement

The problem can be divided into three parts:

1. Displacement 1: 4.0 m at 30° south of west.
2. Displacement 2: 5.0 m east.
3. Displacement 3: 6.0 m at 60° north of east.

To solve this problem, we need to:

1. Find the components of each displacement vector.
2. Sum these components to find the resultant vector.
3. Calculate the magnitude and direction of the resultant vector.

Components of Each Displacement Vector

(1) A particle undergoes three successive displacements in a plane as follows: 4.0 m, 300 south of west; 5.0 m
east; and 6.0 m, 600 north of east.

(a) Find the components of each of the displacement vectors.
(b) Find the components of the resultant displacement.
(c) Calculate the magnitude and direction of the resultant displacement.?

Displacement 1: 4.0 m, 30° South of West

This displacement is directed at an angle of 30° south of west. The direction “south of west” indicates that the vector is pointing to the left (westward) and slightly downward (southward).

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Learn how to calculate the resultant displacement of a particle through vector analysis. This guide covers vector components, magnitude, and direction in physics.

To find the components, we use trigonometry:

• x-component (westward): D1x=4.0 m×cos⁡(30°)D_{1x} = 4.0 \, \text{m} \times \cos(30°)D1x​=4.0m×cos(30°)
• y-component (southward): D1y=4.0 m×sin⁡(30°)D_{1y} = 4.0 \, \text{m} \times \sin(30°)D1y​=4.0m×sin(30°)

Plugging in the values:

• D1x=4.0 m×0.866=3.464 mD_{1x} = 4.0 \, \text{m} \times 0.866 = 3.464 \, \text{m}D1x​=4.0m×0.866=3.464m (westward, so it’s negative in the x-direction)
• D1y=4.0 m×0.5=2.0 mD_{1y} = 4.0 \, \text{m} \times 0.5 = 2.0 \, \text{m}D1y​=4.0m×0.5=2.0m (southward, so it’s negative in the y-direction)

Thus, the components of the first displacement vector are:

• D1x=−3.464 mD_{1x} = -3.464 \, \text{m}D1x​=−3.464m
• D1y=−2.0 mD_{1y} = -2.0 \, \text{m}D1y​=−2.0m

Displacement 2: 5.0 m East

This displacement is purely horizontal, directed toward the east.

• x-component (eastward): D2x=5.0 mD_{2x} = 5.0 \, \text{m}D2x​=5.0m
• y-component: D2y=0.0 mD_{2y} = 0.0 \, \text{m}D2y​=0.0m

Thus, the components of the second displacement vector are:

• D2x=5.0 mD_{2x} = 5.0 \, \text{m}D2x​=5.0m
• D2y=0.0 mD_{2y} = 0.0 \, \text{m}D2y​=0.0m

Displacement 3: 6.0 m, 60° North of East

This displacement is directed at an angle of 60° north of east. The direction “north of east” indicates that the vector is pointing to the right (eastward) and slightly upward (northward).

• x-component (eastward): D3x=6.0 m×cos⁡(60°)D_{3x} = 6.0 \, \text{m} \times \cos(60°)D3x​=6.0m×cos(60°)
• y-component (northward): D3y=6.0 m×sin⁡(60°)D_{3y} = 6.0 \, \text{m} \times \sin(60°)D3y​=6.0m×sin(60°)

Plugging in the values:

• D3x=6.0 m×0.5=3.0 mD_{3x} = 6.0 \, \text{m} \times 0.5 = 3.0 \, \text{m}D3x​=6.0m×0.5=3.0m
• D3y=6.0 m×0.866=5.196 mD_{3y} = 6.0 \, \text{m} \times 0.866 = 5.196 \, \text{m}D3y​=6.0m×0.866=5.196m

Thus, the components of the third displacement vector are:

• D3x=3.0 mD_{3x} = 3.0 \, \text{m}D3x​=3.0m
• D3y=5.196 mD_{3y} = 5.196 \, \text{m}D3y​=5.196m

Resultant Displacement Vector Components

Now that we have the components of each displacement vector, we can sum them to find the components of the resultant displacement vector.

• Resultant x-component: Rx=D1x+D2x+D3xR_x = D_{1x} + D_{2x} + D_{3x}Rx​=D1x​+D2x​+D3x​
• Resultant y-component: Ry=D1y+D2y+D3yR_y = D_{1y} + D_{2y} + D_{3y}Ry​=D1y​+D2y​+D3y​

Substituting the values:

• Rx=−3.464 m+5.0 m+3.0 m=4.536 mR_x = -3.464 \, \text{m} + 5.0 \, \text{m} + 3.0 \, \text{m} = 4.536 \, \text{m}Rx​=−3.464m+5.0m+3.0m=4.536m
• Ry=−2.0 m+0.0 m+5.196 m=3.196 mR_y = -2.0 \, \text{m} + 0.0 \, \text{m} + 5.196 \, \text{m} = 3.196 \, \text{m}Ry​=−2.0m+0.0m+5.196m=3.196m

Thus, the components of the resultant displacement vector are:

• Rx=4.536 mR_x = 4.536 \, \text{m}Rx​=4.536m
• Ry=3.196 mR_y = 3.196 \, \text{m}Ry​=3.196m

Magnitude and Direction of the Resultant Displacement

To find the magnitude of the resultant displacement, we use the Pythagorean theorem:

R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}R=Rx2​+Ry2​​

Substituting the values:

R=(4.536 m)2+(3.196 m)2=20.577+10.212=30.789≈5.55 mR = \sqrt{(4.536 \, \text{m})^2 + (3.196 \, \text{m})^2} = \sqrt{20.577 + 10.212} = \sqrt{30.789} \approx 5.55 \, \text{m}R=(4.536m)2+(3.196m)2​=20.577+10.212​=30.789​≈5.55m

The magnitude of the resultant displacement is approximately 5.55 meters.

To find the direction, we use the inverse tangent function (arctan):

θ=arctan⁡(RyRx)\theta = \arctan\left(\frac{R_y}{R_x}\right)θ=arctan(Rx​Ry​​)

Substituting the values:

θ=arctan⁡(3.196 m4.536 m)≈arctan⁡(0.7044)≈35.26°\theta = \arctan\left(\frac{3.196 \, \text{m}}{4.536 \, \text{m}}\right) \approx \arctan(0.7044) \approx 35.26°θ=arctan(4.536m3.196m​)≈arctan(0.7044)≈35.26°

Thus, the direction of the resultant displacement is approximately 35.26° north of east.

Applications of Vector Analysis in Physics

The analysis of vector components is a fundamental technique in physics, especially in kinematics and dynamics. It allows us to break down complex motion into simpler parts, making it easier to calculate and understand. For instance, when dealing with projectile motion, the initial velocity can be split into horizontal and vertical components, which can then be analyzed independently to determine the trajectory of the projectile.

In this problem, we applied vector analysis to find the resultant displacement of a particle undergoing three successive displacements in a plane. This method is not only applicable to displacement but also to other vector quantities such as velocity, acceleration, and force.

Importance of Understanding Displacement in Physics

Displacement is a key concept in physics because it provides information about the change in position of a particle. Unlike distance, which only measures the total path length traveled, displacement is a vector quantity that takes into account both magnitude and direction. This makes displacement a more comprehensive measure of motion, as it indicates not just how far a particle has moved, but also in which direction.

Understanding how to calculate displacement using vector analysis is crucial for solving a wide range of problems in physics. Whether it’s determining the trajectory of a projectile, analyzing the motion of a car, or understanding the forces acting on an object, displacement plays a central role in understanding the physical world.

Challenges and Common Mistakes in Vector Analysis

While vector analysis is a powerful tool, it can be challenging for students and practitioners alike. Some common mistakes include:

• Incorrect Angle Interpretation: Misinterpreting the angle or direction associated with a vector can lead to incorrect component calculations. It’s important to carefully analyze the direction and use the correct trigonometric functions.
• Sign Errors: Failing to account for the direction of a vector component can result in sign errors. For example, westward and southward components should be negative in the respective coordinate system, while eastward and northward components should be positive.
• Overlooking Units: Always keep track of the units when performing calculations. Inconsistent units can lead to errors in the final result.

To avoid these mistakes, it’s essential to practice problems involving vector analysis regularly and to double-check calculations for accuracy.

Extending the Concept: Displacement in Three Dimensions

While this article focuses on displacements in a two-dimensional plane, the concept can be extended to three dimensions. In three-dimensional space, a displacement vector has three components: $x$, $y$, and $z$. The same principles of vector addition and trigonometry apply, but with an additional dimension to consider.

For example, in three dimensions, the magnitude of a displacement vector is given by:

R=Rx2+Ry2+Rz2R = \sqrt{R_x^2 + R_y^2 + R_z^2}R=Rx2​+Ry2​+Rz2​​

And the direction can be described using spherical coordinates or angles relative to the axes.

Three-dimensional vector analysis is crucial in fields such as engineering, computer graphics, and physics, where objects and particles often move in all three spatial dimensions.

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Conclusion

The problem of a particle undergoing three successive displacements in a plane illustrates the importance of vector analysis in physics. By breaking down each displacement into its components, summing these components, and calculating the resultant vector, we gain a deeper understanding of the particle’s overall motion.

Vector analysis is not only applicable to displacement but also to other vector quantities like velocity, acceleration, and force. Mastering this technique is essential for solving a wide range of problems in physics and engineering.

As we have seen, careful attention to the direction, magnitude, and units of vectors is crucial for accurate calculations. With practice and understanding, vector analysis becomes a powerful tool for analyzing and solving complex problems in physics.

FAQs About Three Successive Displacements in a Plane

1. What is a displacement vector?

   A displacement vector represents the change in position of a particle, including both the magnitude (distance) and direction of the movement. It is a fundamental concept in kinematics.

2. How do you find the components of a displacement vector?

   The components of a displacement vector are found using trigonometric functions. For a vector with a given magnitude and angle, the x-component is calculated using cosine, and the y-component is calculated using sine of the angle.

3. Why is it important to find the resultant displacement?

   The resultant displacement gives the overall change in position after multiple displacements. It provides a single vector that summarizes the combined effect of all individual movements.

4. What is the difference between displacement and distance?

   Displacement is a vector quantity that considers both magnitude and direction, indicating the shortest path between two points. Distance is a scalar quantity that measures the total path length traveled, regardless of direction.

5. Can displacement be negative?

   Yes, displacement can have negative components, depending on the chosen coordinate system. Negative values indicate movement in the opposite direction along the respective axis (e.g., westward or southward in a standard Cartesian plane).