Question

6. I Draw each of the following vectors. Then find its - and components.

a.$\overrightarrow{a}=\left(3.5\text{m}/{\text{s}}^2\right.$, negative x-direction)

b.$\overrightarrow{v}=\left(440\text{m}/\text{s},{30}^{°}\right.$below the positive x-axis )

c.$\overrightarrow{r}=\left(12\text{m},{40}^{°}\right.$ above the positive x-axis )

Short answer

Part (a) x-component and y-component of the vectors are

$\begin{array}{l}\text{x-component}=0\text{m}/{\text{s}}^2\\ \text{y-component}=3.5\text{m}/{\text{s}}^2\end{array}$

Part (b) x-component and y-component of the vectors are

$\begin{array}{l}\text{x-component}=381\text{m}/\text{s}\\ \text{y-component}=-220\text{m}/\text{s.}\end{array}$

Part (c) x-component and y-component of the vectors are
$\begin{array}{l}\text{x-component=}9.19\text{m}\\ \text{y-component}=7.71\text{m}\end{array}$

Step-by-step solution

Given Information

a.$\overrightarrow{a}=\left(3.5\text{m}/{\text{s}}^2\right.$ , negative x -direction)

b. $\overrightarrow{v}=\left(440\text{m}/\text{s},{30}^{°}\right.$below the positive x-axis )

c.$\overrightarrow{r}=\left(12\text{m},{40}^{°}\right.$ above the positive x-axis )

Parts (a)

Vector can be represented as below:

Here,$\theta ={90}^{°}$

x-component is

$\begin{array}{l}{a}_x=-a\cos \theta \\ =-3.5\text{m}/{\text{s}}^2\times \cos {90}^{°}\\ =-3.5\text{m}/{\text{s}}^2\times 0\\ =0\text{m}/{\text{s}}^2\end{array}$

y-component is

$\begin{array}{l}{a}_y=a\sin \theta .\\ =3.5\text{m}/{\text{s}}^2\times \sin {90}^{°}\\ =3.5\text{m}/{\text{s}}^2\times 1\\ =3.5\text{m}/{\text{s}}^2\end{array}$

$\begin{array}{l}\text{x-component}=0\text{m}/{\text{s}}^2\\ \text{y-component}=3.5\text{m}/{\text{s}}^2\end{array}$

Part (b)

Vector$\overrightarrow{v}$can be represented as below:

x-component is

$\begin{array}{l}{v}_x=v\cos \theta \\ =440\text{m}/\text{s}\times \cos {30}^{°}\\ =440\text{m}/\text{s}\times 0.866\\ =381\text{m}/\text{s}\end{array}$

y-component is

$\begin{array}{l}{v}_y=-v\sin \theta \\ =-440\text{m}/\text{s}\times \sin {30}^{°}\\ =-440\text{m}/\text{s}\times 0.5\\ =-220\text{m}/\text{s}.\end{array}$

$\begin{array}{l}\text{x-component}=381\text{m}/\text{s}\\ \text{y-component}=-220\text{m}/\text{s.}\end{array}$

Part (c)

.c. Vector$\overrightarrow{r}$can be represented as below:

x- component is

$\begin{array}{l}{r}_x=r\cos \theta \\ =12\text{m}\times \cos {40}^{°}\\ =12\text{m}\times 0.766\\ =9.19\text{m}\end{array}$

y-component is

$\begin{array}{l}{r}_y=r\sin \theta \\ =12\text{m}\times \sin {40}^{°}\\ =12\text{m}\times 0.643\\ =7.71\text{m}.\\ \text{x-component=}9.19\text{m}\\ \text{y-component}=7.71\text{m}\end{array}$