7. using components, determine the total displacement from the following individual displacements…

Explanation:

Step1: Resolve $\Delta\vec{d}_1$ into components

The west - east component of $\Delta\vec{d}1$ is $d{1x}=- 2.5\sin30.0^{\circ}\text{ m}=-1.25\text{ m}$ (negative for west - direction), and the south - north component is $d_{1y}=-2.5\cos30.0^{\circ}\text{ m}\approx - 2.165\text{ m}$ (negative for south - direction).

Step2: Resolve $\Delta\vec{d}_3$ into components

The west - east component of $\Delta\vec{d}3$ is $d{3x}=4.9\cos38.0^{\circ}\text{ m}\approx3.86\text{ m}$ (positive for east - direction), and the south - north component is $d_{3y}=-4.9\sin38.0^{\circ}\text{ m}\approx - 2.95\text{ m}$ (negative for south - direction).

Step3: Analyze $\Delta\vec{d}_2$ components

The west - east component of $\Delta\vec{d}2$ is $d{2x} = 0\text{ m}$, and the south - north component is $d_{2y}=-3.6\text{ m}$ (negative for south - direction).

Step4: Sum up the x - components

$d_x=d_{1x}+d_{2x}+d_{3x}=-1.25 + 0+3.86 = 2.61\text{ m}$

Step5: Sum up the y - components

$d_y=d_{1y}+d_{2y}+d_{3y}=-2.165-3.6 - 2.95=-8.715\text{ m}$

Step6: Calculate the magnitude of the total displacement

$d=\sqrt{d_x^{2}+d_y^{2}}=\sqrt{(2.61)^{2}+(-8.715)^{2}}\approx9.1\text{ m}$

Step7: Calculate the direction of the total displacement

$\theta=\tan^{-1}\left(\frac{d_y}{d_x}\right)=\tan^{-1}\left(\frac{-8.715}{2.61}\right)\approx - 73.4^{\circ}$ or $73.4^{\circ}$ south of east.

Answer:

The magnitude of the total displacement is approximately $9.1\text{ m}$ in the direction $73.4^{\circ}$ south of east.