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

Step-by-step 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) {(40,0 m)/(-30,0 m) }`

`\theta=233,13\textdegree`

Answer 2

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)((-30)/(50))=126.86^(o)`