91Ó°ÊÓ

The given data can be listed below,

• The x-component of velocity at t=0 is,$v_x=90\mathrm{m}/\mathrm{s}$
• The y-component of velocity at t=0 is, $v_y=110\mathrm{m}/\mathrm{s}$
• The y-component of velocity at t=30 s is,$v_x=-170\mathrm{m}/\mathrm{s}$
• The y-component of velocity at t=30 s is, $v_y=40\mathrm{m}/\mathrm{s}$.

The vector quantity of acceleration describes the change in velocity with respect to time.

The diagram of the velocity vector is given by,

The change in velocity of the plane is given by,

$$\begin{gathered} ∆v=\left(v_{x_2}-v_{x_1}\right)\hat{i}+\left(v_{y_2}-v_{y_1}\right)\hat{j} \\ =\left(-170-90\right)\hat{i}+\left(40-110\right)\hat{j} \\ =\left[-260\hat{i}-70\hat{j}\right]\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

The change in time of the plane is given by,

$$\begin{gathered} ∆t=30-0 \\ =30s \end{gathered}$$

Thus, the change in velocity is $\left[-260\hat{i}-70\hat{j}\right]\frac{\mathrm{m}}{\mathrm{s}}$and the change in time is 30 s .

The average velocity is given by,

$a=\frac{∆v}{∆t}$

Here,$∆v$is the change in velocity and$∆t$is the change in time.

Substitute all the values in the above,

$$\begin{gathered} a=\frac{-260\hat{i}-70\hat{j}}{30} \\ =\left[-8.66\hat{i}-2.33\hat{j}\right]\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

Thus, the average velocity is calculated to be $\left[-8.66\hat{i}-2.33\hat{j}\right]\frac{\mathrm{m}}{{\mathrm{s}}^2}$

The magnitude of acceleration a is given by,

$$\begin{gathered} a=\sqrt{a_x^2+a_y^2} \\ =\sqrt{{\left(-8.66\right)}^2+{\left(-2.33\right)}^2} \\ =8.97\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

The angle$\mathrm{θ}$ is given by,

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{a_y}{a_z}\right) \\ ={\tan }^{-1}\left(\frac{2.33}{8.66}\right) \\ =15° \end{gathered}$$

The acceleration components are of the 3^rd quadrant in the plane, so$180°$ needs to be added to the calculated value $\mathrm{θ}$. So,

$$\begin{gathered} \mathrm{θ}=180°+15° \\ =195° \end{gathered}$$

Thus, the acceleration is$8.97\frac{\mathrm{m}}{{\mathrm{s}}^2}$ and the angle$\mathrm{θ}$ is $195°$