- #1

karush

Gold Member

MHB

- 3,269

- 5

$\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 $-33.15^o$

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