Chapter 3, Problem 3P

If $V_x=\text{9}.\text{8}0$ units and $V_y=-6.40$ units, determine the magnitude and direction of $\text{<mover accent="true"> <mo> V </mo> <mo> → </mo> </mover>}$ .

Solution

The magnitude and direction

Solution: $302.34\text{km}$ and ${13.2}^{\text{o}}$ South of West.

Explanation:

Given:

  $\begin{array}{l}{\text{V}}_{\text{x}}=9.80\text{units}\\ {\text{V}}_y=-6.40\text{units}\end{array}$

Formula used:

The magnitude of the resultant vector and direction can be obtained using the following formulae:

  $\begin{array}{l}R=\sqrt{P^2+Q^2+2PQ\cos \theta }\\ \alpha ={\tan }^{-1}\left(\frac{Q\sin \theta }{P+Q\cos \theta }\right)\end{array}$

Where, P and Q are vectors and $\theta$ is the angle between them.

Calculations:

As per question:

  $\begin{array}{l}P=9.80\\ Q=-6.40\\ \theta ={90}^{\text{o}}\end{array}$

Substitute the values and solve to find the magnitude of the resultant displacement.

  $\begin{array}{l}R=\sqrt{{9.80}^2+{\left(-6.40\right)}^2+2\left(9.80\right)\left(-6.40\right)\cos {90}^{\text{o}}}\\ R=11.7\text{units}\end{array}$

Now, substitute the values to find the direction:

  $\begin{array}{l}\alpha ={\tan }^{-1}\left(\frac{-6.40\sin {90}^{\text{o}}}{9.80-6.40\cos {90}^{\text{o}}}\right)\\ \alpha =-{33.15}^{\text{o}}\end{array}$

It will be directed ${33.15}^{\text{o}}$ below the x -axis.