The displacement is 12.3 m at an angle [tex]168.2^{\circ}[/tex] (measured as counterclockwise from east direction)
Explanation:
The displacement of an object is a vector connecting the initial position to the final position of motion of the object.
In this problem, the motion consists of three parts:
3.0 m east
2.5 m north
15 m west
We define the positive x-axis as the east direction and the positive y-axis as the north direction. Therefore, the components of the displacement along the x- and y- direction are:
[tex]d_x = +3.0 - 15 = -12 m[/tex]
[tex]d_y = +2.5 m[/tex]
Therefore, the magnitude of the displacement is:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(-12.0)^2+(2.5)^2}=12.3 m[/tex]
And the angle is given by:
[tex]\theta = tan^{-1} (\frac{d_y}{d_x})=tan^{-1}(\frac{2.5}{-12.0})=-11.8^{\circ}[/tex]
However, the actual direction is north-west, therefore the real angle is
[tex]\theta=180^{\circ} +(-11.8^{\circ})=168.2^{\circ}[/tex]
Learn more about displacement:
brainly.com/question/3969582
#LearnwithBrainly
