Pessimist Singularitarian

#2
January 4th, 2018,
15:43
Originally Posted by

**karush**
$\tiny{\\ eb 4.d}$

$\textsf{If $v_x=-9.80$ units and $v_y=-6.40$ units,}\\$

$\textit{determine the magnitude and direction of $V$.}$

\begin{align*}\displaystyle

magnitude&=\sqrt{(-9.8)^2+(-6.4)^2}

=\color{red}{11.7047}\\

direction&=\arctan{\left[\frac{6.4}{-9.8}\right]}

=\color{red}{33.15^o}

\end{align*}

ok the direction is actually in the Q4 but $\arctan{\left[\frac{-6.4}{-9.8}\right]}$

is in Q1! Do we just add $180^o$ for that or is it $-33.15^o$

The direction would be in quadrant III, and so recognizing that both components are negative, you would in fact add $\pi$ to the angle:

$ \displaystyle \theta=\pi+\arctan\left(\frac{v_y}{v_x}\right)$

Originally Posted by

**karush**
Also is there symbols for magnitude and direction other than an arrow on the graph

Also where is the latex for degree instead of ^o

I use ^{\circ} for the degree symbol.