Chapter 4, Problem 9P

(a)

To determine

The components of the acceleration of the fish.

Answer to Problem 9P

The horizontal component of acceleration for the fish is $0.800{\text{ m/s}}^2$ and the vertical component of acceleration for the fish is $-0.300{\text{ m/s}}^2$ .

Explanation of Solution

As per given condition in the problem, this case is related to constant acceleration

According to constant acceleration, motion in two dimensions can be modeled as two independent motions in each of the two perpendicular directions associated with the $x$-axes and $y$-axes. That is, any influence in the $y$ direction does not affect the motion in the $x$ direction and vice versa.

At $t=0\text{ s}$, the velocity vector is given as $\overrightarrow{v_i}=\left(4.00\widehat{i}+1.00\widehat{j}\right)\text{ m/s}$ and the position vector is given as $\overrightarrow{r_i}=\left(10.00\widehat{i}-4.00\widehat{j}\right)\text{ m}$.

At $t=20.0\text{ s}$, the final velocity for the fish is given as $\overrightarrow{v_f}=\left(20.0\widehat{i}-5.00\widehat{j}\right)\text{ m/s}$.

Write the general expression for acceleration as.

    $a=\frac{\Delta v}{\Delta t}$

Here, $a$ is the acceleration, $\Delta v$ is the change in velocity and $\Delta t$ is change in time.

Write the expression for acceleration in $x$-axis as.

    $a_x=\frac{v_{fx}-v_{ix}}{\Delta t}$                                                                                                 (I)

Here, $v_{fx}$ is final velocity component of $x$-axis, $v_{ix}$ is the initial velocity component of $x$ -axis and $a_x$ is the acceleration component of $x$-axis.

Write the expression for acceleration in $y$-axis as.

    $a_y=\frac{v_{fy}-v_{iy}}{\Delta t}$                                                                                                 (II)

Here, $v_{fy}$ is final velocity component of $y$-axis, $v_{iy}$ is the initial velocity component of $y$ -axis and $a_y$ is the acceleration component of $y$-axis.

Write the final expression for acceleration in $xy$-plane as.

    $\overrightarrow{a}=a_x\widehat{i}+a_y\widehat{j}$                                                                                      (III)

Here, $\overrightarrow{a}$ is the net acceleration vector.

Conclusion:

Substitute $20.0\text{ m/s}$ for $v_{fx}$ , $4.00\text{ m/s}$ for $v_{ix}$ and $20.0\text{ s}$ for $\Delta t$ in equation (I)    $\begin{array}{c}{a}_x=\frac{20.0\text{ m/s}-4.00\text{ m/s}}{20.0\text{ s}}\\ =\frac{16.0\text{ m/s}}{20.0\text{ s}}\\ =0.800{\text{ m/s}}^2\end{array}$

Substitute $-5.00\text{ m/s}$ for $v_{fy}$ , $1.00\text{ m/s}$ for $v_{iy}$ and $20.0\text{ s}$ for $\Delta t$ in equation (II)

    $\begin{array}{c}{a}_y=\frac{-5.00\text{ m/s}-1.00\text{ m/s}}{20.0\text{ s}}\\ =\frac{-6.00\text{ m/s}}{20.0\text{ s}}\\ =-0.300{\text{ m/s}}^2\end{array}$

Thus, the horizontal component of acceleration for the fish is $0.800{\text{ m/s}}^2$ and the vertical component of acceleration for the fish is $-0.300{\text{ m/s}}^2$ .

(b)

To determine

The direction of the acceleration with respect to unit vector.

Answer to Problem 9P

The direction of the net acceleration is $339°$.

Explanation of Solution

Write the expression for direction of the acceleration of the fish as.

    $\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$                                                                                     (IV)

Here $\theta$ is the direction of velocity components.

Conclusion:

Substitute $-0.3{\text{ m/s}}^2$ for $a_y$ and $0.8{\text{ m/s}}^2$ for $a_x$ in equation (III)

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-0.3{\text{ m/s}}^2}{0.8{\text{ m/s}}^2}\right)\\ ={\tan }^{-1}\left(-0.375\right)\\ =-20.556°\\ \approx -20.6°\end{array}$

The direction of the net acceleration is $360°-20.6°=339.4°\approx 339°$.

Thus, the direction of the net acceleration from the positive $x$-axis is $339°$.

(c)

To determine

The direction and the position of the fish at constant acceleration in $t=25.0\text{ s}$.

Answer to Problem 9P

The $x$ component for the position of the fish at $25.0\text{ s}$ is $360\text{ m}$ , the $y$ component for the position of the fish at $25.0\text{ s}$ is $-72.7\text{ m}$ and the direction of the velocity component is $-15.2°$.

Explanation of Solution

The position vector $\overrightarrow{r_f}$ o a particle is the vector sum of the original position $\overrightarrow{r_i}$, a displacement $\overrightarrow{v_i}\cdot t$ arising from the initial velocity of the particle and a displacement $\frac{1}{2}\overrightarrow{a}{t}^2$ resulting from the constant acceleration of the particle.

Write the expression for final position vector for horizontal component as.

    $x_f=\overrightarrow{x_i}+\overrightarrow{v_{xi}}\cdot t+\frac{1}{2}\overrightarrow{a_x}{t}^2$                                                                              (V)

Here, $x_f$ is final position of horizontal axis, $\overrightarrow{v_{xi}}$ is velocity component at horizontal axis and $\overrightarrow{x_i}$ is original position vector.

Write the expression for final position vector as.

    $y_f=\overrightarrow{y_i}+\overrightarrow{v_{yi}}\cdot t+\frac{1}{2}\overrightarrow{a_y}{t}^2$                                                                                (VI)

Here, $y_f$  is final position of vertical axis, $\overrightarrow{v_{yi}}$ is velocity component at vertical axis and $\overrightarrow{y_i}$ is original position vector.

Write the expression for final position vector as.

    $\overrightarrow{r}=x_f\widehat{i}+y_f\widehat{j}$                                                                                     (VII)

Here, $\overrightarrow{r}$ is resultant position vector.

Write the expression for final velocity at $x$-axis as.

    $v_{xf}=v_{x\widehat{i}}+a_{x}t$                                                                                   (VIII)

Here, $v_{xf}$ is final velocity at horizontal axis.

Write the expression for direction of the velocity components at $25.0\text{ s}$ as.

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

Conclusion:

Substitute $10.0\text{ m}$ for $x_f$, $4.00\text{ m/s}$ for $v_{x\widehat{i}}$, $25.0\text{ s}$ for $t$ and $0.8{\text{ m/s}}^2$ for $a_x$ in equation (V)

    $\begin{array}{c}{x}_f=10.0\text{ m}+\left(4.00\text{ m/s}\right)\left(25.0\text{ s}\right)+\frac{1}{2}\left(0.8\text{ m/s}\right){\left(25.0\text{ s}\right)}^2\\ =10.0\text{ m}+\text{100}\text{.0 m}+\text{250 m}\\ \text{=360 m}\end{array}$

Substitute $-4.00\text{ m}$ for $y_f$, $1.00\text{ m/s}$ for $v_{y\widehat{i}}$, $25.0\text{ s}$ for $t$ and $-0.3{\text{ m/s}}^2$ for $a_y$ in equation (VI)

    $\begin{array}{c}{y}_f=-4.0\text{ m}+\left(1.00\text{ m/s}\right)\cdot \left(25.0\text{ s}\right)+\frac{1}{2}\left(-0.3\text{ m/s}\right){\left(25.0\text{ s}\right)}^2\\ =-4.0\text{ m}+\text{25}\text{.0 m}-\text{93}\text{.75 m}\\ \text{=}-72.75\text{m}\end{array}$

 Thus, the position vector for the fish at $25.0\text{ s}$ is $\left(360\widehat{i}-72.7\widehat{j}\right)\text{ m}$.

Substitute $4.00\text{ m/s}$ for $v_{x\widehat{i}}$, $0.8{\text{ m/s}}^2$ for $a_x$ and $25.0\text{ s}$ for $t$ in equation (VIII)

    $\begin{array}{c}{v}_{xf}=4.00\text{ m/s}+\left(0.8{\text{ m/s}}^2\right)\left(25.0\text{ s}\right)\\ =4.00\text{ m/s}+20.0\text{ m/s}\\ \text{=24 m/s}\end{array}$

Substitute $1.0\text{ m/s}$ for $v_{y\widehat{i}}$, $-0.3{\text{ m/s}}^2$ for $a_y$ and $25.0\text{ s}$ for $t$ in equation (IX)

    $\begin{array}{c}{v}_{yf}=1.0\text{ m/s}+\left(-0.3{\text{ m/s}}^2\right)\left(25.0\text{ s}\right)\\ =1.0\text{ m/s}+\left(-7.5\text{ m/s}\right)\\ =-6.5\text{ m/s}\end{array}$

Substitute $\text{24 m/s}$ for $v_{xf}$ and $-6.5\text{ m/s}$ for $v_{yf}$ in equation (IX)

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-6.5}{24.0}\right)\\ ={\tan }^{-1}\left(-0.2708\right)\\ =-15.2°\end{array}$

Thus, the direction of the velocity component is $-15.2°$.