Q5E - University Physics with Modern Physics

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

Step-by-Step Solution

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

Step 2: Concept/Significance of the acceleration

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

Step 3: (a) Sketch the velocity vectors at t 1 and t 2 . 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,

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

 

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 .

Step 4: (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 $\left[-8.66 \hat{i}-2.33 \hat{j}\right]\frac{\mathrm{m}}{{\mathrm{s}}^2}$ 

Step 5: (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  $195°$