Finding the Displacement Vector

Oct 13, 2025 by orovajewels.com 130 views

Introduction

In physics, understanding the concepts of displacement and force vectors is crucial for analyzing the motion of objects. Displacement vector represents the change in position of an object from its initial to final position, while force vector represents the push or pull on an object. In this article, we will explore how to find the displacement vector and the angle between the force vector and the displacement vector.

Problem Statement

A force $F = 3i + 2j$ pounds is applied to an object that moves from point $(-1, 2)$ to point $(5, 4)$ in feet. We need to find the displacement vector $D$ for the object's movement and the angle between $F$ and $D$ .

Finding the Displacement Vector

The displacement vector $D$ can be found by subtracting the initial position from the final position. Let's denote the initial position as $(-1, 2)$ and the final position as $(5, 4)$ . The displacement vector $D$ can be calculated as:

D = (5 - (-1))i + (4 - 2)j = 6i + 2j

Therefore, the displacement vector $D$ is $6i + 2j$ feet.

Finding the Angle between $F$ and $D$

To find the angle between $F$ and $D$ , we can use the dot product formula:

F \cdot D = |F| |D| \cos \theta

where $|F|$ and $|D|$ are the magnitudes of the force vector and the displacement vector, respectively, and $\theta$ is the angle between them.

First, let's find the magnitudes of the force vector and the displacement vector:

|F| = \sqrt{3^2 + 2^2} = \sqrt{13}

|D| = \sqrt{6^2 + 2^2} = \sqrt{40}

Now, we can plug in the values into the dot product formula:

F \cdot D = (3i + 2j) \cdot (6i + 2j) = 18 + 4 = 22

|F| |D| = \sqrt{13} \sqrt{40} = 2\sqrt{130}

Now, we can solve for $\cos \theta$ :

\cos \theta = \frac{F \cdot D}{|F| |D|} = \frac{22}{2\sqrt{130}} = \frac{11}{\sqrt{130}}

To find the angle $\theta$ , we can take the inverse cosine of both sides:

\theta = \cos^{-1} \left( \frac{11}{\sqrt{130}} \right)

Using a calculator, we can find the value of $\theta$ :

\theta \approx 25.4^\circ

Therefore, the angle between the force vector $F$ and the displacement vector $D$ is approximately $25.4^\circ$ .

Conclusion

$In this article, we have learned how to find the displacement vector and the angle between the force vector and the displacement vector. We have used the concepts of vector addition and the dot product formula to solve the problem. The displacement vector $D$ is $6i + 2j$ feet, and the angle between $F$ and $D$ is approximately $25.4^\circ$.<br/> **Frequently Asked Questions (FAQs) about Displacement and Force Vectors**$

Q: What is the difference between displacement and distance?

A: Displacement is the change in position of an object from its initial to final position, while distance is the total length of the path traveled by the object. For example, if an object moves from point A to point B and then back to point A, the distance traveled is the sum of the distances from A to B and from B to A, but the displacement is zero because the object ends up at the same point where it started.

Q: How do I find the displacement vector?

A: To find the displacement vector, you need to subtract the initial position from the final position. Let's say the initial position is $x_i$ and $y_i$ , and the final position is $x_f$ and $y_f$ . The displacement vector $D$ can be calculated as:

D = (x_f - x_i)i + (y_f - y_i)j

Q: What is the dot product formula?

A: The dot product formula is:

F \cdot D = |F| |D| \cos \theta

where $|F|$ and $|D|$ are the magnitudes of the force vector and the displacement vector, respectively, and $\theta$ is the angle between them.

Q: How do I find the angle between two vectors?

A: To find the angle between two vectors, you can use the dot product formula. First, find the dot product of the two vectors, then divide the result by the product of the magnitudes of the two vectors. The result is the cosine of the angle between the two vectors. To find the angle, take the inverse cosine of the result.

Q: What is the magnitude of a vector?

A: The magnitude of a vector is the length of the vector. It can be calculated as:

|F| = \sqrt{x^2 + y^2}

where $x$ and $y$ are the components of the vector.

Q: Can I use the dot product formula to find the magnitude of a vector?

A: Yes, you can use the dot product formula to find the magnitude of a vector. If you know the dot product of a vector with itself, you can use the formula:

|F|^2 = F \cdot F

to find the magnitude of the vector.

Q: What is the difference between a scalar and a vector?

A: A scalar is a quantity that has only magnitude, while a vector is a quantity that has both magnitude and direction. For example, temperature is a scalar, while velocity is a vector.

Q: Can I add two vectors together?

A: Yes, you can add two vectors together by adding their corresponding components. For example, if you have two vectors $F_1 = x_1i + y_1j$ and $F_2 = x_2i + y_2j$ , the sum of the two vectors is:

F_1 + F_2 = (x_1 + x_2)i + (y_1 + y_2)j

Q: Can I subtract one vector from another?

A: Yes, you can subtract one vector from another by subtracting their corresponding components. For example, if you have two vectors $F_1 = x_1i + y_1j$ and $F_2 = x_2i + y_2j$ , the difference between the two vectors is:

F_1 - F_2 = (x_1 - x_2)i + (y_1 - y_2)j

Q: What is the unit vector?

A: The unit vector is a vector with a magnitude of 1. It is often used to represent the direction of a vector. For example, if you have a vector $F = x_i + y_j$ , the unit vector in the direction of $F$ is:

\hat{F} = \frac{F}{|F|} = \frac{x_i + y_j}{\sqrt{x^2 + y^2}}

Q: Can I use the unit vector to find the angle between two vectors?

A: Yes, you can use the unit vector to find the angle between two vectors. If you know the unit vectors in the direction of the two vectors, you can use the formula:

\cos \theta = \hat{F_1} \cdot \hat{F_2}

to find the cosine of the angle between the two vectors.