Question

A jet plane is flying at a constant altitude. At time${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{}\mathit{s}$, it has components of velocity${\mathit{v}}_{\mathbf{x}}\mathbf{=}\mathbf{90}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$,${\mathit{v}}_{\mathbf{y}}\mathbf{=}\mathbf{110}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. At time${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{30}\mathbf{}\mathit{s}$, the components are${\mathit{v}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{170}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$,${\mathbf{v}}_{\mathbf{y}}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$.

(a) Sketch the velocity vectors at${\mathit{t}}_{\mathbf{1}}$and${\mathit{t}}_{\mathbf{2}}$. How do these two vectors differ? For this time interval calculate

(b) the components of the average acceleration, and

(c) the magnitude and direction of the average acceleration

Short answer

1. 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
2. The average velocity is calculated to be$\left[-8.06\hat{i}-2.33\hat{j}\right]\frac{\mathrm{m}}{{\mathrm{s}}^2}$.
3. The acceleration is$8.97\frac{\mathrm{m}}{{\mathrm{s}}^2}$ and the angle$\theta$ is$195°$

Step-by-step solution

Identification of given data

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, $v_y=40\mathrm{m}/\mathrm{s}$.

Concept/Significance of the acceleration

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

(a) Sketch the velocity vectors at t1 and t2 . How do these two vectors differ?

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 .

(b) Determination of the components of the average acceleration

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 $\left[-8.66\hat{i}-2.33\hat{j}\right]\frac{\mathrm{m}}{{\mathrm{s}}^2}$

(c) Determination of the magnitude and direction of the average acceleration.

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$\theta$ is given by,

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

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

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

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