91Ó°ÊÓ

a.The velocity vector is $\stackrel{→}{v}=\left(7.5\text{m}/\text{s},{30}^{°}\right.$clockwise from the positive y-axis )

b. The acceleration vector is$\stackrel{→}{a}=\left(1.5\text{m}/{\text{s}}^2,{30}^{°}\right.$ above the negative -axis)

c. The force vector is data-custom-editor="chemistry" $\stackrel{→}{F}=\left(50\text{N},{36.9}^{°}\right.$counterclockwise from the positive y-axis )

x- component is

$\begin{array}{l}{v}_{\text{x}}=v\sin \mathrm{θ}\\ =7.5\text{m}/\text{s}×\sin {30}^{°}\\ =7.5\text{m}/\text{s}×0.5\\ =3.75\text{m}/\text{s}\end{array}$

y-component is

$\begin{array}{l}{v}_{\text{y}}=v\cos \mathrm{θ}=7.5\text{m}/\text{s}×\cos {30}^{°}\\ =7.5\text{m}/\text{s}×0.866\\ =6.495\text{m}/\text{s}\\ ≈6.5\text{m}/\text{s}\end{array}$

Vector $\stackrel{→}{a}$can be represented as below:

x-component is

$\begin{array}{l}{a}_x=−a\cos \mathrm{θ}=−1.5\text{m}/{\text{s}}^2×\cos {30}^{°}\\ =−1.5\text{m}/{\text{s}}^2×0.866\\ =−1.299\text{m}/{\text{s}}^2\\ =−1.3\text{m}/{\text{s}}^2\end{array}$

y-component is

$\begin{array}{l}{a}_y=a\sin \mathrm{θ}\\ =1.5\text{m}/{\text{s}}^2×\sin {30}^{°}\\ =1.5\text{m}/{\text{s}}^2×0.5\\ =0.75\text{m}/{\text{s}}^2\end{array}$

Vector $\stackrel{→}{F}$can be represented as below:

x-component is

$\begin{array}{l}{F}_x=−F\sin \mathrm{θ}\\ =−50\text{N}×\sin {36.9}^{°}\\ =−50\text{N}×0.60\\ =−30\text{N}\end{array}$

y-component is

$\begin{array}{l}{F}_y=F\cos \mathrm{θ}\\ =50\text{N}×\cos {36.9}^{°}\\ =50\text{N}×0.799\\ =39.98\text{N}\\ ≈40\text{N}\end{array}$