Chapter 8.7, Problem 43HP

To determine

To calculate:

The resultant velocity and direction of $\stackrel{\rightharpoonup }{x}$

Answer to Problem 43HP

The resultant velocity and direction of $\stackrel{\rightharpoonup }{x}$ is $26.6°$ and $4.47m/s$ respectively.

Explanation of Solution

Given information:

  

Calculation:

Draw a line as shown in below figure:

  

Now, use the definition of the tangent to get value of resultant direction in $\stackrel{\rightharpoonup }{x}$

:

  

  $\begin{array}{l}\tan \alpha =\frac{2}{4}\\ \alpha ={\tan }^{-1}\left(\frac{2}{4}\right)\\ \alpha =26.6°\end{array}$

Use the Pythagoras theorem to get value of resultant velocity in $\stackrel{\rightharpoonup }{x}$ :

  $\begin{array}{l}\stackrel{\rightharpoonup }{x}=\sqrt{2^2+4^2}\\ \stackrel{\rightharpoonup }{x}=\sqrt{20}\\ \stackrel{\rightharpoonup }{x}=4.47m/s\end{array}$