How To Find The Net Displacement

Finding the net displacement is a fundamental concept in physics, particularly in the study of motion. Whether you're dealing with simple linear movements or more complex multi-dimensional scenarios, understanding how to calculate displacement is crucial. Displacement is a vector quantity that refers to the shortest distance from the initial to the final position of an object. It differs from distance, which measures the total length of the path traveled. This article provides a comprehensive guide on how to find the net displacement, covering various methods, practical examples, and frequently asked questions to ensure a thorough understanding.

Introduction

Imagine a runner completing a lap on a 400-meter track. Although the runner has covered a distance of 400 meters, their net displacement is zero because they end up back where they started. This simple example illustrates the difference between distance and displacement. Displacement is concerned only with the change in position, not the path taken. Understanding displacement is essential in numerous applications, from navigation and engineering to sports science and computer simulations. Let's delve into the methods and techniques for accurately determining net displacement.

Understanding Displacement

Definition and Significance

Displacement is defined as the change in position of an object. It is a vector quantity, meaning it has both magnitude (size) and direction. The magnitude of displacement is the shortest distance between the initial and final positions, and the direction points from the initial to the final position.

The significance of displacement lies in its ability to simplify the analysis of motion. Instead of tracking every twist and turn of an object's path, we can focus on the start and end points. This simplification is particularly useful in complex scenarios where the exact path is either unknown or irrelevant.

Key Differences: Distance vs. Displacement

Feature	Distance	Displacement
Definition	Total length of the path traveled	Shortest distance between initial and final positions
Type	Scalar quantity (magnitude only)	Vector quantity (magnitude and direction)
Consideration	Considers the entire path taken	Considers only the initial and final positions
Value	Always positive or zero	Can be positive, negative, or zero
Example	Running a 400-meter track: Distance = 400 meters	Running a 400-meter track: Displacement = 0 meters

Basic Formula for Displacement

The most basic formula for displacement is:

[ \Delta x = x_f - x_i ]

Where:

(\Delta x) is the displacement,
(x_f) is the final position, and
(x_i) is the initial position.

This formula applies to one-dimensional motion, such as movement along a straight line.

Methods to Calculate Net Displacement

Several methods can be used to calculate net displacement, depending on the nature of the motion and the available information.

1. One-Dimensional Motion

In one-dimensional motion, the object moves along a straight line. The displacement can be positive or negative, indicating the direction of movement.

Example: A car starts at position (x_i = 2) meters and moves to a final position of (x_f = 10) meters. The displacement is:

[ \Delta x = 10 , \text{m} - 2 , \text{m} = 8 , \text{m} ]

The positive value indicates that the car moved in the positive direction (e.g., to the right).

2. Two-Dimensional Motion

Two-dimensional motion involves movement in a plane, requiring the use of vector components to determine the net displacement.

Vector Components Method

Step 1: Break down each displacement into its x and y components. If an object moves with a displacement (d) at an angle (\theta) from the horizontal, the components are: [ d_x = d \cos(\theta) ] [ d_y = d \sin(\theta) ]
Step 2: Sum the x components and the y components separately. [ \Delta x_{\text{net}} = \sum d_x ] [ \Delta y_{\text{net}} = \sum d_y ]
Step 3: Use the Pythagorean theorem to find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(\Delta x_{\text{net}})^2 + (\Delta y_{\text{net}})^2} ]
Step 4: Determine the direction of the net displacement using the inverse tangent function. [ \theta = \arctan\left(\frac{\Delta y_{\text{net}}}{\Delta x_{\text{net}}}\right) ]

Example: An object undergoes two displacements:

(d_1 = 5) meters at an angle of (30^\circ)
(d_2 = 8) meters at an angle of (120^\circ)

Step 1: Find the x and y components of each displacement. [ d_{1x} = 5 \cos(30^\circ) \approx 4.33 , \text{m} ] [ d_{1y} = 5 \sin(30^\circ) = 2.5 , \text{m} ] [ d_{2x} = 8 \cos(120^\circ) = -4 , \text{m} ] [ d_{2y} = 8 \sin(120^\circ) \approx 6.93 , \text{m} ]
Step 2: Sum the x and y components. [ \Delta x_{\text{net}} = 4.33 , \text{m} + (-4) , \text{m} = 0.33 , \text{m} ] [ \Delta y_{\text{net}} = 2.5 , \text{m} + 6.93 , \text{m} = 9.43 , \text{m} ]
Step 3: Find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(0.33 , \text{m})^2 + (9.43 , \text{m})^2} \approx 9.43 , \text{m} ]
Step 4: Find the direction of the net displacement. [ \theta = \arctan\left(\frac{9.43}{0.33}\right) \approx 87.99^\circ ]

Thus, the net displacement is approximately (9.43) meters at an angle of (87.99^\circ) from the horizontal.

Graphical Method (Vector Addition)

Step 1: Draw each displacement vector to scale, tip-to-tail. Start with the first displacement vector, then place the tail of the second vector at the tip of the first, and so on.
Step 2: Draw the resultant vector from the tail of the first vector to the tip of the last vector. This resultant vector represents the net displacement.
Step 3: Measure the length and direction of the resultant vector. Use a ruler to measure the length (magnitude) and a protractor to measure the angle (direction).

Example: Consider the same two displacements as before:

(d_1 = 5) meters at an angle of (30^\circ)
(d_2 = 8) meters at an angle of (120^\circ)

Draw a vector of 5 meters at 30 degrees, then draw a vector of 8 meters at 120 degrees starting from the tip of the first vector. The resultant vector can be measured to find its magnitude and direction, which should approximately match the values calculated using the vector components method.

3. Three-Dimensional Motion

Three-dimensional motion extends the concepts of two-dimensional motion into three dimensions, requiring an additional z-component.

Vector Components Method

Step 1: Break down each displacement into its x, y, and z components. [ d_x = d \sin(\theta) \cos(\phi) ] [ d_y = d \sin(\theta) \sin(\phi) ] [ d_z = d \cos(\theta) ] Where (\theta) is the polar angle and (\phi) is the azimuthal angle.
Step 2: Sum the x, y, and z components separately. [ \Delta x_{\text{net}} = \sum d_x ] [ \Delta y_{\text{net}} = \sum d_y ] [ \Delta z_{\text{net}} = \sum d_z ]
Step 3: Use the three-dimensional Pythagorean theorem to find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(\Delta x_{\text{net}})^2 + (\Delta y_{\text{net}})^2 + (\Delta z_{\text{net}})^2} ]
Step 4: Determine the direction of the net displacement using appropriate trigonometric functions. [ \theta = \arccos\left(\frac{\Delta z_{\text{net}}}{|\Delta d|}\right) ] [ \phi = \arctan\left(\frac{\Delta y_{\text{net}}}{\Delta x_{\text{net}}}\right) ]

Example: An object undergoes two displacements:

(d_1 = 4) meters at (\theta_1 = 60^\circ) and (\phi_1 = 30^\circ)
(d_2 = 6) meters at (\theta_2 = 45^\circ) and (\phi_2 = 120^\circ)

Step 1: Find the x, y, and z components of each displacement. [ d_{1x} = 4 \sin(60^\circ) \cos(30^\circ) = 3 , \text{m} ] [ d_{1y} = 4 \sin(60^\circ) \sin(30^\circ) \approx 1.73 , \text{m} ] [ d_{1z} = 4 \cos(60^\circ) = 2 , \text{m} ] [ d_{2x} = 6 \sin(45^\circ) \cos(120^\circ) \approx -2.12 , \text{m} ] [ d_{2y} = 6 \sin(45^\circ) \sin(120^\circ) \approx 3.67 , \text{m} ] [ d_{2z} = 6 \cos(45^\circ) \approx 4.24 , \text{m} ]
Step 2: Sum the x, y, and z components. [ \Delta x_{\text{net}} = 3 , \text{m} + (-2.12) , \text{m} = 0.88 , \text{m} ] [ \Delta y_{\text{net}} = 1.73 , \text{m} + 3.67 , \text{m} = 5.4 , \text{m} ] [ \Delta z_{\text{net}} = 2 , \text{m} + 4.24 , \text{m} = 6.24 , \text{m} ]
Step 3: Find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(0.88 , \text{m})^2 + (5.4 , \text{m})^2 + (6.24 , \text{m})^2} \approx 8.31 , \text{m} ]
Step 4: Find the direction of the net displacement. [ \theta = \arccos\left(\frac{6.24}{8.31}\right) \approx 41.4^\circ ] [ \phi = \arctan\left(\frac{5.4}{0.88}\right) \approx 80.7^\circ ]

Thus, the net displacement is approximately (8.31) meters at (\theta \approx 41.4^\circ) and (\phi \approx 80.7^\circ).

Practical Examples

Example 1: A Hiker’s Journey

A hiker walks 5 km east, then 3 km north, and finally 2 km west. Find the hiker's net displacement.

Step 1: Break down the displacements into components.
- East: (d_{1x} = 5 , \text{km}), (d_{1y} = 0 , \text{km})
- North: (d_{2x} = 0 , \text{km}), (d_{2y} = 3 , \text{km})
- West: (d_{3x} = -2 , \text{km}), (d_{3y} = 0 , \text{km})
Step 2: Sum the components. [ \Delta x_{\text{net}} = 5 , \text{km} + 0 , \text{km} + (-2) , \text{km} = 3 , \text{km} ] [ \Delta y_{\text{net}} = 0 , \text{km} + 3 , \text{km} + 0 , \text{km} = 3 , \text{km} ]
Step 3: Find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(3 , \text{km})^2 + (3 , \text{km})^2} = \sqrt{18} \approx 4.24 , \text{km} ]
Step 4: Find the direction of the net displacement. [ \theta = \arctan\left(\frac{3}{3}\right) = 45^\circ ]

The hiker's net displacement is approximately (4.24) km at an angle of (45^\circ) north of east.

Example 2: An Airplane Flight

An airplane flies 200 km at (60^\circ) north of east, then 300 km due south. What is the plane's net displacement?

Step 1: Break down the displacements into components. [ d_{1x} = 200 \cos(60^\circ) = 100 , \text{km} ] [ d_{1y} = 200 \sin(60^\circ) \approx 173.2 , \text{km} ] [ d_{2x} = 0 , \text{km} ] [ d_{2y} = -300 , \text{km} ]
Step 2: Sum the components. [ \Delta x_{\text{net}} = 100 , \text{km} + 0 , \text{km} = 100 , \text{km} ] [ \Delta y_{\text{net}} = 173.2 , \text{km} + (-300) , \text{km} = -126.8 , \text{km} ]
Step 3: Find the magnitude of the net displacement. [ |\Delta d| = \sqrt{(100 , \text{km})^2 + (-126.8 , \text{km})^2} \approx 161.55 , \text{km} ]
Step 4: Find the direction of the net displacement. [ \theta = \arctan\left(\frac{-126.8}{100}\right) \approx -51.7^\circ ]

The airplane's net displacement is approximately (161.55) km at an angle of (51.7^\circ) south of east.

Common Mistakes to Avoid

Confusing Distance and Displacement: Always remember that displacement is not the same as distance. Displacement is the shortest path between the initial and final positions, while distance is the total path length.
Incorrectly Applying Vector Components: Ensure that you correctly resolve vectors into their components using trigonometric functions. Pay attention to the angles and their reference points.
Ignoring Direction: Displacement is a vector quantity, so direction is crucial. Always include the direction in your final answer.
Unit Inconsistencies: Make sure all measurements are in the same units before performing calculations. Convert units as needed.
Rounding Errors: Avoid rounding intermediate results too early, as this can lead to significant errors in the final answer.

FAQ (Frequently Asked Questions)

Q: Can displacement be zero even if the object moved? A: Yes, displacement can be zero if the object returns to its initial position, regardless of the distance traveled.

Q: Is displacement a scalar or vector quantity? A: Displacement is a vector quantity, meaning it has both magnitude and direction.

Q: How do you find the direction of displacement? A: The direction of displacement is found using trigonometric functions, such as the inverse tangent function, based on the components of the displacement vector.

Q: What is the significance of negative displacement? A: A negative displacement indicates that the object moved in the negative direction relative to the chosen coordinate system.

Q: Can displacement be greater than distance? A: No, the magnitude of displacement can never be greater than the distance traveled. It can be equal to the distance if the motion is along a straight line in one direction.

Conclusion

Finding net displacement is a crucial skill in physics and engineering, enabling the analysis of motion by focusing on the change in position rather than the entire path traveled. Whether dealing with one-dimensional, two-dimensional, or three-dimensional motion, understanding the methods of vector components and graphical addition is essential. By breaking down complex movements into manageable components and applying basic trigonometric principles, one can accurately determine both the magnitude and direction of net displacement. Remember to avoid common mistakes such as confusing distance with displacement and neglecting to account for direction.

How do you plan to apply these principles to solve real-world problems involving motion and displacement? What other strategies do you find helpful in mastering vector calculations?