Q6E - University Physics with Modern Physics

Question

A dog running in an open field has components of velocity ${\mathbf{v}}_{\mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ and ${\mathbf{v}}_{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$ at ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{ }\mathit{s}$. For the time interval from ${\mathit{t}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{ }\mathit{s}$ to ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{ }\mathit{s}$, the average acceleration of the dog has magnitude $\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{ }\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$ and direction 31.0° measured from the +x-axis toward the +y-axis. At ${\mathit{t}}_{\mathbf{2}}\mathbf{=}\mathbf{20}\mathbf{ }\mathit{s}$, (a) what are the x- and y-components of the dog’s velocity? (b) What are the magnitude and direction of the dog’s velocity? (c) Sketch the velocity vectors at ${\mathit{t}}_{\mathbf{1}}$ and ${\mathit{t}}_{\mathbf{2}}$. How do these two vectors differ?

Step-by-Step Solution

Answer

1. The x and y-component of the velocity of dog at $t=20 s$ is $6.46 \mathrm{m}/\mathrm{s}$ and $0.52 \mathrm{m}/\mathrm{s}$ respectively.
2. The magnitude of the velocity of the dog is 6.48 m/s and direction of dog is $4.6°$.
3. The magnitude of velocity and direction of dog at time $t=10 \mathrm{s}$ and $t=20 \mathrm{s}$ is different.

Step 1: Identification of given data

The given data can be listed below,

• The x-component of velocity at $t_1=10 \mathrm{s}$ is ${\mathrm{v}}_{\mathrm{x}}=2.6 \mathrm{m}/\mathrm{s}$.
• The y-component of velocity at $t_1=10 \mathrm{s}$ is ${\mathrm{v}}_{\mathrm{y}}=-1.8 \mathrm{m}/\mathrm{s}$.
• The time interval 10 s is 20 s . 
• The average acceleration of dog is $a_{avg}=0.45 \mathrm{m}/{\mathrm{s}}^2$ .
• The direction of dog is $31°$.

Step 2: Concept/Significance of average acceleration

Acceleration is the rate at which velocity changes. A change in either the magnitude or the direction of velocity, which is a vector variable with both magnitude and direction, denotes that the moving body is experiencing an acceleration

Step 3: Determination of the x- and y-components of the dog’s velocity.

(a)

 

The component form of acceleration of the dog is given by,

$a=a_x \hat{i}+a_y \hat{j}$

 

The x-component of the acceleration is given by,

${\left(a_{avg}\right)}_x=a_{avg}\cos \theta$

Substitute $0.45 \mathrm{m}/{\mathrm{s}}^2$ for ${\mathrm{a}}_{\mathrm{avg}}$ and $31°$ for $\theta$ in the above equation.

 

$$\begin{gathered} {\left(a_{avg}\right)}_x=0.45\cos 31° \\ =0.386 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

The y-component of the acceleration is given by,

 

Substitute $0.45 \mathrm{m}/{\mathrm{s}}^2$ for $a_{avg}$ and $31°$ for $\theta$ in the above equation.

 

$$\begin{gathered} {\left(a_{avg}\right)}_y=0.45 \sin 31° \\ =0.232 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

 

The average acceleration of the dog is given by,

$a_{avg}=\frac{v_2-v_1}{t_2-t_1}$ 

 

Here, $v_2$ is the final velocity of dog at time 20 s, is the initial velocity of dog, and $t_2-t_1$ is the time interval.

 

The x-component velocity of the dog is given by,

${\left(v_x\right)}_2={\left(v_x\right)}_1+{\left(a_{avg}\right)}_x\left(t_2-t_1\right)$

 

Substitute 2.6 m/s for ${\left(v_x\right)}_1$, $0.386\mathrm{m}/{\mathrm{s}}^2$ for ${\left(a_{avg}\right)}_x$, 20 s for $t_2$ and $10s$ for $t_1$ in the above equation.

 

$$\begin{gathered} {\left(v_x\right)}_1=2.6 \mathrm{m}/\mathrm{s}+0.386 \mathrm{m}/\mathrm{s}\left(20 \mathrm{s}-10 \mathrm{s}\right) \\ =6.46 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

The y-component velocity of the dog is given by,

${\left(v_y\right)}_2={\left(v_y\right)}_1+\left(a_{avg}\right)\left(t_2-t_1\right)$

 

Substitute $1.8 \mathrm{m}/\mathrm{s}$ for ${\left(v_y\right)}_1$, $0.232\mathrm{m}/{\mathrm{s}}^2$ for ${\left(a_{avg}\right)}_y$ , 20s for $t_2$ and 10 s for $t_1$ in the above equation.

 

$$\begin{gathered} {\left(v_y\right)}_2=1.8 \mathrm{m}/\mathrm{s}+0.232\left(20 \mathrm{s}-10 \mathrm{s}\right) \\ =0.52 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, the x and y-component of the velocity of dog at $t=20 s$ is $6.46 \mathrm{m}/\mathrm{s}$ and $0.52 \mathrm{m}/\mathrm{s}$ respectively.

Step 4: Determination of the magnitude and direction of the dog’s velocity at time

(b)

 

The magnitude of velocity of the dog is given by,

$\left|v\right|=\sqrt{v_x^2+v_y^2}$ 

 

Substitute all the values in the above,

$$\begin{gathered} v=\sqrt{{\left(6.46 \mathrm{m}/\mathrm{s}\right)}^2+{\left(0.52 \mathrm{m}/\mathrm{s}\right)}^2} \\ =6.48 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

The direction of dog at $t=20 s$ is given by,

$$\begin{gathered} \tan \alpha =\frac{v_y}{v_x} \\ \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \end{gathered}$$

 

Substitute 0.52m/s for $v_y$ and $6.48\mathrm{m}/\mathrm{s}$ for $v_x$ in the above equation.

$$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \\ ={\tan }^{-1}\left(\frac{0.52}{6.48}\right) \\ =4.6° \end{gathered}$$ 

 

Thus, the magnitude of the velocity of the dog is 6.48 m/s and direction of dog is $4.6°$.

Step 5: Sketch of the the velocity vectors at and .

(c)

 

The magnitude of velocity of dog at time $t=10 s$ is given by,

$\left|v\right|=\sqrt{{\left(v_x\right)}_1^2+{\left(v_y\right)}_1^2}$

 

Substitute 2.6m/s for ${\left(v_x\right)}_1$ and $-1.8\mathrm{m}/\mathrm{s}$ for ${\left(v_y\right)}_1$ in the above equation.

 

$$\begin{gathered} v=\sqrt{{\left(2.6 \mathrm{m}/\mathrm{s}\right)}^2+{\left(-1.8 \mathrm{m}/\mathrm{s}\right)}^2} \\ =3.16 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

The direction of dog at time $t=10 \mathrm{s}$ is,

 $$\begin{gathered} \alpha ={\tan }^{-1}\left(\frac{v_y}{v_x}\right) \\ ={\tan }^{-1}\left(\frac{-1.8}{2.6}\right) \\ =325.3° \end{gathered}$$

 

The diagram for the velocity vectors is shown below,

 

                       

 

Thus, the magnitude of velocity and direction of dog at time $t=10 s$ and $t=20 s$ is different.