Question

displacement in 2 - dimensions
directions: for each scenario, calculate the displacement. you will need to use the diagram to visualize the movement and to determine how many meters galaxy is moving.
remember: your answer should have direction and magnitude!
pythagorean theorem: $a^{2}+b^{2}=c^{2}\to c = sqrt{a^{2}+b^{2}}$

1. galaxy walks from her house to inspect the tree again, then she catches the scent of some stinky shrubs and walks south to smell them.

what is her displacement from home when she is at the shrubs?
show work!

1. from home, galaxy walks to inspect a pile of trash some hooligan threw in the backyard.

what is her displacement?
show work!

1. from home, galaxy runs to catch a field mouse.

what is her displacement?
show work!

Explanation:

Since no diagram is provided, we'll assume a general 2 - dimensional coordinate system with the starting - point (home) as the origin \((0,0)\) and solve the problems conceptually.

Problem 6

Step1: Identify the components

Assume the first movement (to the tree) is along the \(x\) - axis (say \(x\) meters) and the second movement (south) is along the \(y\) - axis (say \(y\) meters in the negative \(y\) - direction).
Let the displacement along the \(x\) - axis be \(x\) and along the \(y\) - axis be \(-y\).

Step2: Calculate the magnitude

Using the Pythagorean theorem \(d=\sqrt{x^{2}+y^{2}}\), where \(d\) is the magnitude of the displacement.

Step3: Determine the direction

The direction \(\theta\) can be found using \(\tan\theta=\frac{-y}{x}\), and the direction is described as \(\theta\) degrees south of east.

However, without values of \(x\) and \(y\), we can only give the general formula for the displacement. The magnitude of the displacement \(d = \sqrt{x^{2}+y^{2}}\) and the direction is \(\arctan(\frac{-y}{x})\) measured from the positive \(x\) - axis (south of east).

Problem 7

Step1: Identify the components

Let the displacement vector from the origin (home) to the trash - pile have components \(x\) and \(y\) in the \(x\) and \(y\) directions respectively.

Step2: Calculate the magnitude

Using the Pythagorean theorem \(d=\sqrt{x^{2}+y^{2}}\), where \(d\) is the magnitude of the displacement.

Step3: Determine the direction

The direction \(\theta=\arctan(\frac{y}{x})\) (adjust the angle according to the quadrant in which the end - point lies).

Problem 8

Step1: Identify the components

Let the displacement vector from the origin (home) to the position of the field - mouse have components \(x\) and \(y\) in the \(x\) and \(y\) directions respectively.

Step2: Calculate the magnitude

Using the Pythagorean theorem \(d = \sqrt{x^{2}+y^{2}}\), where \(d\) is the magnitude of the displacement.

Step3: Determine the direction

The direction \(\theta=\arctan(\frac{y}{x})\) (adjust the angle according to the quadrant in which the end - point lies).

Since no numerical values are given in the problem, we cannot provide a numerical answer. But the general method to find the displacement (magnitude and direction) for each problem is as described above.

Answer:

For problem 6: Magnitude \(d=\sqrt{x^{2}+y^{2}}\), direction \(\arctan(\frac{-y}{x})\) south of east.
For problem 7: Magnitude \(d=\sqrt{x^{2}+y^{2}}\), direction \(\arctan(\frac{y}{x})\) (adjusted for the quadrant).
For problem 8: Magnitude \(d=\sqrt{x^{2}+y^{2}}\), direction \(\arctan(\frac{y}{x})\) (adjusted for the quadrant).