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?

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 $31°$.

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

(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 $t=20s$is $6.46\mathrm{m}/\mathrm{s}$and$0.52\mathrm{m}/\mathrm{s}$ respectively.

(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$t=20s$ 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 $4.6°$.

(c)

The magnitude of velocity of dog at time$t=10s$ 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$t=10s$ and$t=20s$ is different.