Chapter 4, Problem 70P

(a)

To determine

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos.

Answer to Problem 70P

The distance travelled by Carlos and Shannon when Shannon catches up to Carlos is $\underset{\_}{77\text{m}}$.

Explanation of Solution

The free body diagram of the given situation is shown in the Figure 1.

Let the positive x direction be down the slope and the positive y direction is in the direction of normal force.

Since there is no force in the vertical direction, the net force in the vertical direction is zero.

    $n=mg\cos \theta$        (I)

Here, $n$ is the normal force, $m$ is the mass, $g$ is the acceleration due to gravity, and $\theta$ is the angle makes the slope with horizontal.

From the free body diagram, write the expression for net force in the horizontal direction.

    $mg\sin \theta -f_k=m{a}_x$        (II)

Here, $f_k$ is the frictional force, and $a_x$ is the acceleration in the horizontal direction.

Use ${\mu }_{k}n$ for $f_k$, equation (I) in (II), and solve for $a_x$.

    $\begin{array}{c}mg\sin \theta -{\mu }_{k}mg\cos \theta =m{a}_x\\ a_x=g\left(\sin \theta -{\mu }_k\cos \theta \right)\end{array}$        (III)

Write the kinematic equation to find the Carole’s velocity after travelling the $5\text{m}$ distance.

    $v_{{}_{f,C}}^2=v_{i,C}^2+2a_{x,C}\Delta x$        (IV)

Here, $v_{f,C}$ is the Carlos’s final velocity, $v_{i,C}$ is the Carlos’s initial velocity, $a_{x,C}$ is the Carlos’s initial acceleration, and $\Delta x$ is the distance.

Write the kinematic equation to find the distance travelled by Carlos at the beginning pf the time interval.

    $x_{f,C}=x_{i,C}+v_{i,C}t+\frac{1}{2}{a}_{x,C}{t}^2$        (V)

Here, $x_{f,C}$ is the final position of the Carlos, $x_{i,C}$ is the initial position of the Carlos, and $t$ is the time taken by Shannon to reach Carlos.

Write the kinematic equation to find the distance travelled by Carlos at the beginning of the time interval.

    $x_{f,S}=x_{i,S}+v_{i,S}t+\frac{1}{2}{a}_{x,S}{t}^2$        (VI)

Here, $x_{f,S}$ is the final position of the Shannon, and $x_{i,S}$ is the initial position of the Shannon.

Conclusion:

Substitute, $9.8\text{m}/{\text{s}}^2$ for $g$, $12°$ for $\theta$, and $0.1$ for ${\mu }_k$ in the equation (III) to find acceleration of Carlos.

    $\begin{array}{c}{a}_{x,C}=9.8\text{m}/{\text{s}}^2\left(\sin 12°-0.1\cos 12°\right)\\ =1.07894\text{m}/{\text{s}}^2\end{array}$

Substitute, $9.8\text{m}/{\text{s}}^2$ for $g$, $12°$ for $\theta$, and $0.01$ for ${\mu }_k$ in the equation (III) to find acceleration of Shannon.

    $\begin{array}{c}{a}_{x,C}=9.8\text{m}/{\text{s}}^2\left(\sin 12°-0.01\cos 12°\right)\\ =1.94167\text{m}/{\text{s}}^2\end{array}$

Substitute, $0$ for $v_{i,C}$, $5\text{m}$ for $\Delta x$, and $1.07894\text{m}/{\text{s}}^2$ for $a_{x,C}$ in the equation (IV), and solve for $v_{f,C}$.

    $\begin{array}{c}{v}_{{}_{f,C}}^2=0+2\left(1.07894\text{m}/{\text{s}}^2\right)\left(5\text{m}\right)\\ v_{f,C}=3.28\text{m}/\text{s}\end{array}$

Substitute, $5\text{m}$ for $x_{i,C}$, $3.28\text{m}/\text{s}$ for $v_{i,C}$, and $1.07894\text{m}/{\text{s}}^2$ for $a_{x,C}$ in the equation (V), to find $x_{f,C}$.

        $\begin{array}{c}{x}_{f,C}=5\text{m}+\left(3.28\text{m}/\text{s}\right)t+\frac{1}{2}\left(1.07894\text{m}/{\text{s}}^2\right)t^2\\ =5\text{m}+\left(3.28\text{m}/\text{s}\right)t+\left(0.53947\text{m}/{\text{s}}^2\right)t^2\end{array}$        (VII)

Substitute, $0$ for $x_{i,S}$, $0$ for $v_{i,S}$, and $1.94167\text{m}/{\text{s}}^2$ for $a_{x,S}$ in the equation (VI), to find $x_{f,S}$.

        $\begin{array}{c}{x}_{f,S}=0+0+\frac{1}{2}\left(84167\text{m}/{\text{s}}^2\right)t^2\\ =\text{0}\text{.970}\text{m}/{\text{s}}^2{t}^2\end{array}$        (VIII)

The distance travelled by the Shannon and Carlos are same.

Equate equation (VII) and (VIII), and solve for $t$.

    $\begin{array}{c}5\text{m}+\left(3.28\text{m}/\text{s}\right)t+\left(0.53947\text{m}/{\text{s}}^2\right)t^2=\text{0}\text{.970}\text{m}/{\text{s}}^2{t}^2\\ \left(0.4313\text{m}/{\text{s}}^2\right)t^2-\left(3.28\text{m}/\text{s}\right)t-5\text{m}=0\\ t=8.9\text{s}\end{array}$

Substitute, $8.9\text{s}$ for $t$ in the equation (VIII), to find $x_{f,S}$.

    $\begin{array}{c}{x}_{f,S}=\text{0}\text{.970}\text{m}/{\text{s}}^2\times {\left(8.9\text{s}\right)}^2\\ =77\text{m}\end{array}$

Therefore, the distance travelled by Carlos and Shannon when Shannon catches up to Carlos is $\underset{\_}{77\text{m}}$.

(b)

To determine

The speed of the Shannon with respect to Carlos.

Answer to Problem 70P

The speed of the Shannon with respect to Carlos is $\underset{\_}{4.4\text{m}/\text{s}}$.

Explanation of Solution

Write the kinematic equation.

    $v=u+at$        (IX)

Conclusion:

Substitute, $0$ for $u$, $1.94167\text{m}/{\text{s}}^2$ for $a$, and $8.9\text{s}$ for $t$ in the equation (IX), to find the velocity of Shannon after time $t$ $\left(v_{f,S}\right)$.

    $\begin{array}{c}{v}_{f,S}=0+\left(1.94167\text{m}/{\text{s}}^2\right)\left(8.9\text{s}\right)\\ =17.28\text{m}/\text{s}\end{array}$

Substitute, $3.28\text{m}/{\text{s}}^2$ for $u$, $1.07894\text{m}/{\text{s}}^2$ for $a$, and $8.9\text{s}$ for $t$ in the equation (IX), to find the velocity of Carlos after time $t$ $\left(v_{f,C}\right)$.

    $\begin{array}{c}{v}_{f,C}=3.28\text{m}/{\text{s}}^2+\left(1.07894\text{m}/{\text{s}}^2\right)\left(8.9\text{s}\right)\\ =12.88\text{m}/\text{s}\end{array}$

Since Carlos is moving in the same direction as Shannon, the relative speed of Shannon with respect to Carlos can be calculated as follows,

    $\begin{array}{c}{v}_{SC}=17.28\text{m}/\text{s}-12.88\text{m}/\text{s}\\ =4.4\text{m}/\text{s}\end{array}$

Therefore, the speed of the Shannon with respect to Carlos is $\underset{\_}{4.4\text{m}/\text{s}}$.