If$V_x=9.80\mathrm{units}$and$V_y=-6.40\mathrm{units}$, determine the magnitude and direction of$\overrightarrow{V}$.

The magnitude of a vector represents its length graphically.

The magnitude of the X-component of $\overrightarrow{V}$,$V_{\mathrm{x}}=9.80\mathrm{units}$

The magnitude of the Y-component of $\overrightarrow{V}$, $V_{\mathrm{y}}=-6.40\mathrm{units}$

Let the magnitude of $\overrightarrow{V}$be V.

For two vectors $\overrightarrow{A}$and $\overrightarrow{B}$, oriented at angle$\theta$ with respect to each other, the magnitude of their resultant vector or their vector sum is given by the formula

$R=\sqrt{A^2+B^2+2AB\cos \theta }$.

Here, A and B are magnitudes of vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, respectively, andR is the magnitude of the resultant vector.

In this problem, $V_{\mathrm{x}}$and $V_{\mathrm{y}}$are the rectangular components of $\overrightarrow{V}$. The angle between them is $90°$.

Using the formula for magnitude of vector sum, you get

$\begin{array}{c}V=\sqrt{V_{\mathrm{x}}^2+V_{\mathrm{y}}^2+2V_{\mathrm{x}}{V}_{\mathrm{y}}\cos 90°}\\ =\sqrt{{9.80}^2+{6.40}^2}\\ =11.70\mathrm{units}.\end{array}$

Thus, the magnitude of $\overrightarrow{V}$is 11.70 units.

The tangent function of angle $\theta$can be written as $\tan \theta =\frac{V_{\mathrm{y}}}{V_{\mathrm{x}}}$.

Substituting the values,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-6.40}{9.80}\right)\\ =-33.15°.\end{array}$

Thus, $\overrightarrow{V}$is oriented at an angle of $-33.15°$with respect to the positive direction of the X-axis. Here, the negative sign indicates that the angle is measured clockwise from the positive X-axis.