Chapter 39, Problem 2PQ

(a)

To determine

The distance travel by the Desmond in $30.0\text{s}$, with respect to farmer.

Answer to Problem 2PQ

The distance travel by the Desmond in $30.0\text{s}$ is $2.29\times {10}^3\text{m}$.

Explanation of Solution

The speed of Desmond’s train is towards east.

Write the formula for distance traveled in time $t$.

    $d=vt$                                                                                   (I)

Here, $d$ is distance, $v$ is velocity and $t$ is time.

Find the relative velocity of Desmond with respect to farmer.

    $\begin{array}{l}{v}_{\text{DF}}=v_D-0\\ v_{\text{DF}}=v_D\end{array}$

Here, $v_{\text{DF}}$ the velocity of Desmond with respect to farmer and $v_D$ is the velocity of Desmond’s train.

Substitute $v_{\text{D}}$ for $v$ in equation (I) to find $d$.

    $d=v_{\text{D}}t$                                                                              (II)

Conclusion:

Substitute $275\text{km}/\text{h}$ for $v_{\text{D}}$ and $30.0\text{s}$ for $t$ in equation (II) to find $d$.

    $\begin{array}{c}d=\left(275.0\text{km}/\text{h}\right)30.0\text{s}\\ =\left(275.0\text{km}/\text{h}\times \frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)30.0\text{s}\\ =2291.67\text{m}\\ =2.29\times {10}^3\text{m}\end{array}$

Therefore, distance travel by the Desmond in $30.0\text{s}$ is $2.29\times {10}^3\text{m}$.

(b)

To determine

The distance travel by the Lilani in $30.0\text{s}$, with respect to farmer.

Answer to Problem 2PQ

The distance travel by the Lilani in $30.0\text{s}$ is $2.54\times {10}^3\text{m}$.

Explanation of Solution

Lilani will travel with speed of train with respect to the farmer because farmer is stationary.

Find the velocity of Desmond with respect to farmer.

    $\begin{array}{l}{v}_{\text{LF}}=v_L-0\\ v_{\text{LF}}=v_L\end{array}$

Here, $v_{\text{LF}}$ the velocity of Lilani with respect to farmer and $v_L$ is the velocity of Lilani’s train.

Substitute $v_{\text{L}}$ for $v$ in equation (I) to find $d$.

    $d=v_{\text{L}}t$                                                                                                    (III)

Conclusion:

Substitute $305\text{km}/\text{h}$ for $v_{\text{L}}$ and $30.0\text{s}$ for $t$ in equation (III) to find distance.

    $\begin{array}{c}d=\left(305.0\text{km}/\text{h}\right)30.0\text{s}\\ =\left(305.0\text{km}/\text{h}\left(\frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\right)30.0\text{s}\\ =2541.66\text{m}\\ =2.54\times {10}^3\text{m}\end{array}$

Therefore, distance travel by the Lilani in $30.0\text{s}$ is $2.54\times {10}^3\text{m}$.

(c)

To determine

The distance travelled by Desmond with respect to Lilani.

Answer to Problem 2PQ

The distance travel by the Desmond with respect to Lilani in $30.0\text{s}$ is $-250\text{m}$.

Explanation of Solution

Calculate the relative velocity of Desmond with respect to Lilani.

    $v_{\text{L}}=v_{\text{D}}+v_{\text{rel}}$

Here, $v_{\text{rel}}$ is relative velocity of the Desmond with respect to Lilani and $v_{\text{L}}$ is velocity of the Lilani.

Rewrite the above relation for $v_{\text{rel}}$

$\begin{array}{c}{v}_{\text{rel}}=v_{\text{L}}-v_{\text{D}}\\ =-275\text{km}/\text{h}-305.0\text{km}/\text{h}\\ =-5.80\times {10}^2\text{km}/\text{h}\end{array}$

Write the formula for distance traveled in time $t$.

    $d=v_{\text{rel}}t$ (IV)

Conclusion:

Substitute $-5.80\times {10}^2\text{km}/\text{h}$ for $v_{\text{rel}}$ and $30.0\text{s}$ for $t$ in equation (IV) to find distance.

    $\begin{array}{c}d=\left(-5.80\times {10}^2\text{km}/\text{h}\right)\left(30.0\text{s}\right)\\ =\left(-5.80\times {10}^2\text{km}/\text{h}\times \frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\left(30.0\text{s}\right)\\ =-4.83\times {10}^3\text{m}\\ \left|d\right|=4.83\times {10}^3\text{m}\end{array}$

The negative sign indicates that Desmond lags behind $250\text{m}$ in $30\text{s}$ than Lilani.

Therefore, distance travel by the Desmond with respect to Lilani in $30.0\text{s}$ is $4.83\times {10}^3\text{m}$.

(d)

To determine

The distance travelled by Lilani with respect to Desmond.

Answer to Problem 2PQ

The distance travel by Lilani with respect to Desmond in $30.0\text{s}$ is $4.83\times {10}^3\text{m}$.

Explanation of Solution

Calculate the relative velocity of Lilani with respect to Desmond.

    $\begin{array}{c}{v}_{\text{rel}}=v_{\text{L}}-v_{\text{D}}\\ =-275\text{km}/\text{h}-305.0\text{km}/\text{h}\\ =-5.80\times {10}^2\text{km}/\text{h}\end{array}$

Here, $v_{\text{rel}}$ is relative velocity of Lilani with respect to Desmond.

Write the formula for distance traveled in $t$ time.

    $d=v_{\text{rel}}t$                                                                                     (V)

Conclusion:

Substitute $-5.80\times {10}^2\text{km}/\text{h}$ for $v_{\text{rel}}$ and $30.0\text{s}$ for $t$ in equation (V) to find distance.

    $\begin{array}{c}d=\left(-5.80\times {10}^2\text{km}/\text{h}\right)\left(30.0\text{s}\right)\\ =\left(-5.80\times {10}^2\text{km}/\text{h}\times \frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\left(30.0\text{s}\right)\\ =-4.83\times {10}^3\text{m}\\ \left|d\right|=4.83\times {10}^3\text{m}\end{array}$

The positive value indicates that Lilani goes $250\text{m}$ ahead in $30\text{s}$ than Desmond.

Therefore, distance travel by Lilani with respect to Desmond in $30.0\text{s}$ is $4.83\times {10}^3\text{m}$.