Vector overrightarrow (v)_(1) is 6.6 units long and points along the nega- tive x axis . Vector overrightarrow (V)_(2) is 8.5 units long and points at +55^circ to the positive x axis. (a)What are the x and y components of each vector'? (b ) Determine the sum overrightarrow (V)_(1)+overrightarrow (V)_(2) (magnitude and angle).

To solve this problem, we need to break down each vector into its x and y components using trigonometric functions. Then, we'll sum the vectors by adding their respective components. Finally, we'll calculate the magnitude and angle of the resultant vector.<br /><br />### (a) Finding the x and y components of each vector<br /><br />#### Vector \(\overrightarrow{v}_1\):<br />- Magnitude: 6.6 units<br />- Direction: Along the negative x-axis (which is \(-180^\circ\) or \(180^\circ\))<br /><br />The x and y components are:<br />\[ \overrightarrow{v}_1 = (6.6 \cos 180^\circ, 6.6 \sin 180^\circ) \]<br />\[ \overrightarrow{v}_1 = (-6.6, 0) \]<br /><br />#### Vector \(\overrightarrow{V}_2\):<br />- Magnitude: 8.5 units<br />- Direction: \(+55^\circ\) to the positive x-axis<br /><br />The x and y components are:<br />\[ \overrightarrow{V}_2 = (8.5 \cos 55^\circ, 8.5 \sin 55^\circ) \]<br />Using a calculator:<br />\[ \cos 55^\circ \approx 0.5736 \]<br />\[ \sin 55^\circ \approx 0.8192 \]<br /><br />So,<br />\[ \overrightarrow{V}_2 = (8.5 \times 0.5736, 8.5 \times 0.8192) \]<br />\[ \overrightarrow{V}_2 \approx (4.85, 6.96) \]<br /><br />### (b) Determining the sum \(\overrightarrow{v}_1 + \overrightarrow{V}_2\)<br /><br />To find the sum, we add the corresponding components of the vectors:<br />\[ \overrightarrow{v}_1 + \overrightarrow{V}_2 = (-6.6 + 4.85, 0 + 6.96) \]<br />\[ \overrightarrow{v}_1 + \overrightarrow{V}_2 = (-1.75, 6.96) \]<br /><br />Now, we calculate the magnitude of the resultant vector:<br />\[ |\overrightarrow{v}_1 + \overrightarrow{V}_2| = \sqrt{(-1.75)^2 + 6.96^2} \]<br />\[ |\overrightarrow{v}_1 + \overrightarrow{V}_2| = \sqrt{3.0625 + 48.4416} \]<br />\[ |\overrightarrow{v}_1 + \overrightarrow{V}_2| = \sqrt{51.5041} \]<br />\[ |\overrightarrow{v}_1 + \overrightarrow{V}_2| \approx 7.19 \]<br /><br />Next, we find the direction (angle) of the resultant vector:<br />\[ \theta = \tan^{-1}\left(\frac{6.96}{-1.75}\right) \]<br />\[ \theta = \tan^{-1}\left(-3.97\right) \]<br />\[ \theta \approx -75.5^\circ \]<br /><br />Since the angle is negative and we need it to be positive, we add \(180^\circ\) to get the direction relative to the positive x-axis:<br />\[ \theta \approx 180^\circ - 75.5^\circ \]<br />\[ \theta \approx 104.5^\circ \]<br /><br />### Summary:<br />(a) The x and y components are:<br />- \(\overrightarrow{v}_1 = (-6.6, 0)\)<br />- \(\overrightarrow{V}_2 = (4.85, 6.96)\)<br /><br />(b) The sum \(\overrightarrow{v}_1 + \overrightarrow{V}_2\) has:<br />- Magnitude: \(\approx 7.19\) units<br />- Angle: \(\approx 104.5^\circ\) to the positive x-axis