Question

A particle is initially at xi = 3m, yi = −5m, and after a while it is found at the coordinates xf = −4m, yf = 2m. (a) On the grid below (next page), draw the initial and final position vectors, and the displacement vector. (b) What are the components of the displacement vector? (c) What are the magnitude and direction of the displacement vector? (You can specify the direction by the angle it makes with either the positive x or the positive y axis.)

Answer 1

Answer:

a) see attachement.

b) x = -7 y=7

c) $M= 2\sqrt{7}$ and $\theta=45° from the y axis.$

Step-by-step explanation:

To draw a vector, first draw a point in the position you want to by looking at the x axis and the y axis. Then, draw a line between the origin (0, 0) and the point you just plotted, and draw an arrow indicating the direction of the vector.

In the plot shown, the blue arrow is the starting position, the green is the final position and the red is the displacement vector.

To calculate the components just operate each component (the components are x and y) of the position vectors:

The bar over x and y indicates that the result inside parenthesis is in this direction.

$\bar{d}=\bar{p_f}-\bar{p_i}=(x_f-x_i)\bar{x}+(y_f-y_i)\bar{y}=(-4-3)\bar{x}+(2-(-5))\bar{y}=(-7)\bar{x}+(7)\bar{y}$

The magnitud can be calculated using the next formula:

$M = \sqrt{x^2+y^2}$

$M=\sqrt{(-7)^2+(7)^2}=\sqrt{2(7)^2}=7\sqrt{2}$

oTo calculate the direction, center the vector we found in the origin (see second graphic):

The positive y axis is the one going up, therefore we use the arc tangent function to calculate it:

$\theta=arctan(\frac{x}{y} )=arctan(\frac{7}{7} )= 45°$ from the y axis