Q7E - University Physics with Modern Physics

Question

C The coordinates of a bird flying in the -plane are given by $\mathit{x}\left(t\right)\mathbf{=}\mathit{\alpha }\mathit{t}$ and $\mathit{y}\left(t\right)\mathbf{=}\mathbf{3}\mathbf{ }\mathit{m}\mathbf{-}\mathit{\beta }{\mathit{t}}^{\mathbf{2}}$, where $\mathit{\alpha }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{ }\mathit{m}\mathbf{/}\mathit{s}$ and $\mathit{\beta }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ .(a) Sketch the path of the bird between $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{s}$ and $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{s}$. (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird’s velocity and acceleration at $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{s}$. (d) Sketch the velocity and acceleration vectors at $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{ }\mathit{s}$ . At this instant, is the bird’s speed increasing, decreasing, or not changing? Is the bird turning? If so, in what direction?

Step-by-Step Solution

Answer

1. Based on the above calculation graph for the path of bird is shown below as,
2. The velocity vector and acceleration vector are $\alpha \hat{i}-2\beta t \hat{j}$ and $-2\beta \hat{j}$.
3. The magnitude and direction of velocity is 5.37 m/s and $296.56°$ respectively and the magnitude of acceleration is $2.4 \mathrm{m}/{\mathrm{s}}^2$ and direction in the negative y-direction.
4. The bird is moving faster and turning in y-direction.

Step 1: Identification of given data

The given data can be listed below,

• The x coordinate of the bird is $x\left(t\right)=\alpha t$. 
• The y-coordinate of the bird is $y\left(t\right)=3m-\beta t^2$. 
• The value of a constant is $\alpha =2.4 \mathrm{m}/\mathrm{s}$. 
• The value of constant for y-coordinate is $\beta =1.2 \mathrm{m}/{\mathrm{s}}^2$.

Step 2: Concept/Significance of average velocity.

An object's average velocity is calculated by dividing its entire displacement by its total travel duration. It is, in other words, the pace at which an item shifts from one location to another.

Step 3: Sketch of the path of the bird between t = 0   s and t = 2   s .

(a)

 

The x-component of displacement is given by,

$x=\alpha t$ 

 

Substitute 2.4 m/s for $\alpha$ in the above equation.

 

$x=\left(2.4 \mathrm{m}/\mathrm{s}\right)t$ 

 

The displacement at $t=0 \mathrm{s}$ 

$$\begin{gathered} x=2.4 \mathrm{m}/\mathrm{s}\left(0\right) \\ =0 \end{gathered}$$ 

 

The displacement at $t=1 s$ is calculated as,

 $$\begin{gathered} \mathrm{x}=\left(2.4 \mathrm{m}/\mathrm{s}\right)\left(1 \mathrm{s}\right) \\ =2.4 \mathrm{m} \end{gathered}$$

 

The displacement at $t=2 s$ is calculated as,

 $$\begin{gathered} \mathrm{x}=\left(2.4 \mathrm{m}/\mathrm{s}\right)\left(2 \mathrm{s}\right) \\ =4.8 \mathrm{m} \end{gathered}$$

 

The y-component of the displacement is given by,

 $y=3 \mathrm{m}-\beta t^2$

 

Substitute $1.2\mathrm{m}/{\mathrm{s}}^2$ for $\beta$ in the above equation.

$y=3 \mathrm{m}-\left(1.2 \mathrm{m}/{\mathrm{s}}^2\right)t^2$

 

The displacement at  $t=0 \mathrm{s}$ is given by,

$$\begin{gathered} y=3 \mathrm{m}-\left(1.2 \mathrm{m}/{\mathrm{s}}^2\right){\left(0 \mathrm{s}\right)}^2 \\ =3 \mathrm{m} \end{gathered}$$

 

The displacement at $t=1 \mathrm{s}$ is given by,

$$\begin{gathered} y=3 \mathrm{m}-\left(1.2 \mathrm{m}/{\mathrm{s}}^2\right){\left(1 \mathrm{s}\right)}^2 \\ =1.8 \mathrm{m} \end{gathered}$$

The displacement at $t=2 s$ is given by,

$$\begin{gathered} y=3 \mathrm{m}-\left(1.2 \mathrm{m}/{\mathrm{s}}^2\right){\left(1 \mathrm{s}\right)}^2 \\ =-1.8 \mathrm{m} \end{gathered}$$

 

Thus, based on the above calculation graph for the path of bird is shown below as,

                                        

Step 4: Determination of the velocity and acceleration vectors of the bird as functions of time.

(b)

 

According to the definition, instantaneous velocity of the bird is given by,

$v=\frac{dx}{dt}$

Here, x is the displacement of bird.

 

Substitute $\alpha t$ for $x$ in the above equation.

 

$$\begin{gathered} v_x=\frac{d\left(\alpha t\right)}{dt} \\ =\alpha \\ =2.4 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

The value of y-component of velocity is given by,

 

$$\begin{gathered} v_y=\frac{d\left(3 m-\beta t^2\right)}{dt} \\ =-2\beta t \\ =-2.4t \end{gathered}$$ 

 

The velocity vector is given by,

$\overrightarrow{v}=\alpha \hat{i}-2 \beta t\hat{j}$

 

The acceleration of the bird is given by,

$a=\frac{dv}{dt}$ 

Here, v is the velocity of the bird.

 

Substitute the value of velocity in the above expression for x-component of acceleration is given by,

$$\begin{gathered} a_x=\frac{d\alpha }{dt} \\ =0 \end{gathered}$$

 

The y-component of acceleration is given by,

$$\begin{gathered} a_y=\frac{d\left(-2 \beta t\right)}{dt} \\ =-2 \beta \\ =-2.4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

Thus, the velocity vector and acceleration vector are $\alpha \hat{i}-2\beta t\hat{j}$ and $-2\beta \hat{j}$.

Step 5: Determination of the magnitude and direction of the bird’s velocity and acceleration at t = 2   s .

(c)

 

The magnitude of the velocity of the bird is given by,

$v=\sqrt{v_x^2+v_y^2}$ 

Here, $v_x$ is the x-component of the velocity, and $v_y$ is the y-component of the velocity.

 

The velocity of the bird at $t=2 s$ is given by,

${\overrightarrow{v}}_{t=2}=\left(2.4 m/s\right)\hat{i}-2\left(2.4 m/s\right)\hat{j}$

 

Substitute 2.4 m/s for $v_x$ and $-4.8 \mathrm{m}/\mathrm{s}$ for  $v_y$ in the above equation.

 

$$\begin{gathered} v=\sqrt{{\left(2.4 \mathrm{m}/\mathrm{s}\right)}^2+{\left(-4.8 \mathrm{m}/\mathrm{s}\right)}^2} \\ =5.37 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

The direction of the velocity of bird is given by,

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$

 

Substitute 2.4 m/s for $v_x$ and $-4.8 \mathrm{m}/\mathrm{s}$ for $v_y$ in the above equation.

 

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{-4.8}{2.4}\right) \\ =-63.44° \\ =296.56° \end{gathered}$$

 

The acceleration at $t=2 s$ is given by,

$$\begin{gathered} {\overrightarrow{a}}_{t=2}=-2.4 \hat{j} \\ =\left(-2.4 \mathrm{m}/{\mathrm{s}}^2\right) \hat{j} \end{gathered}$$

 

The magnitude of acceleration is given by,

 $\left|a\right|=2.4 \mathrm{m}/{\mathrm{s}}^2$

 

Thus, the magnitude and direction of velocity is 5.37 m/s and $296.56°$ respectively and the magnitude of acceleration is $2.4 \mathrm{m}/{\mathrm{s}}^2$ and direction in the negative y-direction.

Step 6: Sketch the velocity and acceleration vectors at t = 2 s . At this instant, is the bird’s speed increasing, decreasing, or not changing.

(d)

 

The graph of the velocity and acceleration at t = 2 s of the bird is shown below,

                                                                     

                                                                               

 

The graph depicts the acceleration vector as having two halves, one parallel to the velocity and the other perpendicular. The parallel component and the velocity component both point in the same direction, indicating that the velocity is rising. The bird will adjust its trajectory in the y direction due to the perpendicular component.

Thus, the bird is moving faster and turning in y-direction.