91Ó°ÊÓ

This law tells us that if the two vectors are represented as the two sides of a triangle, then the third side of the triangle will be the resultant vector.

The resultant side of the triangle gives the distance the dog needs to run to the south of her original starting point.

The given data can be listed below as:

• The dog runs to a distance of $12.0m$in the east.
• The dog then runs to a distance of $28.0m$to $50.0°$west of north.
• The dog will end up at $10.0m$south of where she started.

The free body diagram of the resultant vector can be expressed as:

Analyzing the above figure. It can be stated as:

$$\begin{gathered} \stackrel{I}{R}=\stackrel{I}{A}+\stackrel{I}{B}+\stackrel{I}{C} \\ \stackrel{r}{C}=\stackrel{r}{R}-\stackrel{r}{A}-\stackrel{r}{B} \end{gathered}$$

Here$\stackrel{I}{R}$is the resultant vector in which the dos will end up at south of where she started,$\stackrel{I}{C}$is the velocity the dog needs to run at the east of south, $\stackrel{I}{A}$is the distance traveled by the dog in the east, and$\stackrel{I}{B}$is the distance traveled by the dog in the west of north.

After the addition of the component on the x-axis, we get,

$$\begin{gathered} C_X=R_X-A_X-B_X \\ =0-A-\left(-B\sin 50°\right) \\ =0-12m+\left(28m\right)\sin 50° \\ =9.4m \end{gathered}$$

Here $\stackrel{I}{R}$is the resultant vector in which the dos will end up at south of where she started on the x-axis, $\stackrel{I}{C}$is the required vector in the x-axis, $\stackrel{I}{A}$is the distance traveled by the dog in the east in the x-axis, and $\stackrel{I}{B}$ is the distance traveled by the dog in the west of north in the x-axis.

After the addition of the component on the y-axis, we get,

$$\begin{gathered} C_y=R_y-A_Y-B_Y \\ =-R-0-\left(B\cos 50°\right) \\ =10m-0-\left(28m\right)\cos 50° \\ =27.9m \end{gathered}$$

Here $\stackrel{I}{R}$ is the resultant vector in which the dos will end up at south of where she started in the y axis, $\stackrel{I}{C}$ is the required vector in the y axis, $\stackrel{I}{A}$is the distance traveled by the dog in the east in the y axis, and $\stackrel{I}{B}$ is the distance traveled by the dog in the west of north in the y axis.

Hence, the velocity the dog needs to run at the east of south is expressed as,

$$\begin{gathered} C=\sqrt{{\left(C_x\right)}^2+{\left(C_y\right)}^2} \\ =\sqrt{{\left(9.4m\right)}^2+{\left(-27.9m\right)}^2} \\ =29.6m \end{gathered}$$

The angle direction the dog has to run is:

$$\begin{gathered} \mathrm{Î\pm }={\tan }^{-1}\left(\frac{9.4m}{27.9m}\right) \\ =18.6° \end{gathered}$$

Thus, the dog has to run $29.6m$at the east of south at an angle of $18.6°$from her original starting point.