Question

The x and y components of vector S are -30.0 m and +40.0 m, respectively. Find the magnitude of S and the angle between the direction of S and the positive direction of the x axis. Draw the vector on a two axis coordinate system as well.

Answer 1

Answer:

The given vector can be represented in unit vector as

$\overrightarrow{w}=-30\widehat{i}+40\widehat{j}$

The magnitude of any vector $\overrightarrow{r}=u\widehat{i}+v\widehat{j}$ is given by

$|w|=\sqrt{u^{2}+v^{2}}$

Applying values we get

$|w|=\sqrt{-30^{2}+40^{2}}\\\\|r|=50$

We know that positive x axis in vertorial form is represented as

$\overrightarrow{r}=\widehat{i}$

taking dot product of both the vector's we get

$\overrightarrow{r}.\overrightarrow{w}=|r||w|cos(\theta )\\\\\therefore cos(\theta )=\frac{(-30\widehat{i}+40\widehat{j}).\widehat{i}}{50}\\\\\therefore \theta =cos^{-1}(\frac{-30}{50})=126.86^{o}$

Answer 2

Answer: $||S||=50,0m$, $\theta=233,13\textdegree$

Explanation:

Let $S=(-30.0 m, +40,0 m)$, the magnitude is given by the following formula:

$||S|| = \sqrt{(-30,0m)^{2}+(40,0m)^{2}}$

$||S||=50,0m$

The angle between the direction of $S$ and the positive direction of the x axis is:

$\theta = 180\textdegree- \tan^{-1} {\frac{40,0 m}{-30,0 m} }$

$\theta=233,13\textdegree$