Question

A disoriented physics professor drives 3.25 km north, then 2.20 km west, and then 1.50 km south. Find the magnitude and direction of the resultant displacement, using the method of components. In a vector-addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained by using the method of components.

Short answer

Resultant displacement is 2.81 km at 38.4° north of west.

Step-by-step solution

Identify the components

First, identify the components of each displacement vector. The professor's movements can be broken down into segments based on the cardinal directions. Movement north is positive in the y-direction, west in the negative x-direction, and south in the negative y-direction.

Calculate individual components

For the northward displacement (3.25 km), the components are: \(x_1 = 0\) km and \(y_1 = 3.25\) km. For the westward movement (2.20 km), the components are: \(x_2 = -2.20\) km and \(y_2 = 0\) km. For the southward movement (1.50 km), the components become: \(x_3 = 0\) km and \(y_3 = -1.50\) km.

Sum the components

Add up the x-components and y-components separately to determine the total x and y components of the resultant vector. \[x_{total} = x_1 + x_2 + x_3 = 0 + (-2.20) + 0 = -2.20\,\text{km} \]\[y_{total} = y_1 + y_2 + y_3 = 3.25 + 0 + (-1.50) = 1.75\,\text{km}\]

Calculate the magnitude of the resultant displacement

Use the Pythagorean theorem to find the magnitude of the resultant displacement vector:\[R = \sqrt{x_{total}^2 + y_{total}^2} = \sqrt{(-2.20)^2 + (1.75)^2} = \sqrt{4.84 + 3.06} = \sqrt{7.90}\,\text{km} \approx 2.81\,\text{km}\]

Determine the direction of the resultant vector

Calculate the angle \( \theta \) relative to the west direction using the tangent function:\[\theta = \tan^{-1}\left(\frac{y_{total}}{|x_{total}|}\right) = \tan^{-1}\left(\frac{1.75}{2.20}\right) \approx 38.4^\circ\]Thus, the direction is 38.4° north of west.

Verify with a vector-addition diagram

Illustrate the vectors on a diagram. Draw the 3.25 km north vector, then the 2.20 km west vector, followed by the 1.50 km south vector. Draw the resultant vector from the origin of the first vector to the end of the last vector. The magnitude and direction from the diagram should qualitatively match the calculated 2.81 km at 38.4° north of west.

Key concepts

Displacement Components

When dealing with vector addition, breaking vectors into their components is a fundamental step. Displacement vectors, like any other vectors, have both magnitude and direction. The direction can be resolved into two perpendicular components: typically, these are the x (horizontal) and y (vertical) directions. For our exercise, the professor's movement comprises three segments, each contributing to the overall displacement.

• The northward movement (3.25 km) has components: x-component = 0 km and y-component = 3.25 km.
• The westward movement (2.20 km) has components: x-component = -2.20 km and y-component = 0 km.
• The southward movement (1.50 km) has components: x-component = 0 km and y-component = -1.50 km.

Combining these individual components into total x and y components allows for further calculation and analysis.

Magnitude Calculation

The magnitude of the resultant vector can be interpreted as the total "length" of the displacement, which gives us a sense of how far the physics professor ended up from their starting point. To find it, we use the Pythagorean theorem, applicable due to the orthogonal nature of our component axes:\[ R = \sqrt{x_{ ext{total}}^2 + y_{ ext{total}}^2} \]Plug in our calculated total components:\[ R = \sqrt{(-2.20)^2 + (1.75)^2} = \sqrt{4.84 + 3.06} = \sqrt{7.90} \approx 2.81 \, \text{km} \]Thus, the magnitude of the professor’s resultant displacement is approximately 2.81 km.

Direction Determination

Determining the direction of a vector is just as critical as finding its magnitude, as direction gives the vector its true character. To find the direction of the resultant displacement, we calculate the angle using the inverse tangent function (or arctan). This angle gives the direction relative to the westward direction:\[ \theta = \tan^{-1}\left(\frac{y_{ ext{total}}}{|x_{ ext{total}}|}\right) = \tan^{-1}\left(\frac{1.75}{2.20}\right) \approx 38.4^\circ \text{ north of west} \]Here, the calculation tells us the resultant displacement is 38.4° north of the west direction. This direction determination completes our understanding of the vector's orientation.

Vector Diagram

A vector diagram provides a visual representation to support our calculations and offers a clear illustration of how vectors combine. In this diagram: - Start by drawing the first vector 3.25 km north. - Add the second vector 2.20 km west, starting from the head of the first vector. - Then draw the last vector 1.50 km south from the head of the second vector. - Finally, draw the resultant vector starting from the origin of the first vector to the endpoint of the last vector. This graphical method reflects both the magnitude and direction we've calculated, showing qualitative agreement with the calculated 2.81 km at 38.4° north of west. Seeing the vectors placed end to end emphasizes the precision and correctness of vector addition through graphical means.