Chapter 3, Problem 48P

(a)

To determine

The change in velocity during the trip.

Answer to Problem 48P

The change in velocity during the entire trip is $\left(\text{160}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}$.

Explanation of Solution

Write the expression for change in velocity

    $\Delta \overrightarrow{v}={\overrightarrow{v}}_{\text{final}}-{\overrightarrow{v}}_{\text{initial}}$        (I)

Here, $\Delta \overrightarrow{v}$ is the change in velocity, ${\overrightarrow{v}}_{\text{final}}$ is the final velocity and ${\overrightarrow{v}}_{\text{initial}}$ is the initial velocity.

Write the expression for initial velocity

    ${\overrightarrow{v}}_{\text{final}}=\left(v_3\sin \theta \right)\widehat{\text{i}}-\left(v_3\cos \theta \right)\widehat{\text{j}}$        (II)

Here, $v_3$ is the speed of the final velocity vector.

Conclusion:

Substitute $100\text{ km /h}$ for $v_3$ and $45°$ for $\theta$ in (II) to find ${\overrightarrow{v}}_{\text{final}}$

    $\begin{array}{c}{\overrightarrow{v}}_{\text{final}}=100\text{ km/h}\left(\sin 45°\right)\widehat{\text{i}}-100\text{ km/h}\left(\cos 45°\right)\widehat{\text{j}}\\ =\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}\end{array}$

Substitute $-90\text{ km/h }\widehat{\text{i}}$ for ${\overrightarrow{v}}_{\text{initial}}$ and $\left(7\text{0}\text{.7 km}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km}\right)\widehat{\text{j}}$ for ${\overrightarrow{v}}_{\text{final}}$ in (I) to find $\Delta \overrightarrow{v}$

    $\begin{array}{c}\Delta \overrightarrow{v}=\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}-\left(-90\text{ km/h}\right)\widehat{\text{i}}\\ =\left(\text{160}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}\end{array}$

Thus, the change in velocity during the entire trip is $\left(\text{160}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}$.

(b)

To determine

The average acceleration during the trip.

Answer to Problem 48P

The average acceleration during the entire trip is $\left(260\widehat{\text{i}}+114.5\widehat{\text{j}}\right){\text{km/h}}^2$.

Explanation of Solution

Write the expression for average acceleration

    ${\overrightarrow{a}}_{\text{avg}}=\frac{\Delta \overrightarrow{v}}{\Delta t}$        (III)

Here, ${\overrightarrow{a}}_{\text{avg}}$ is the average acceleration and $\Delta t$ is the time interval.

Write the expression for total time of the trip

    $\Delta t=\frac{d_1}{v_1}+\frac{d_2}{v_2}+\frac{d_3}{v_3}$        (IV)

Here, $d_1,d_2,d_3$ are the distance travelled and $t_1,t_2,t_3$ are times of travel.

Conclusion:

Substitute $16\text{ km}$ for $d_1$, $90\text{ km/h}$ for $v_1$, $8.0\text{ km}$ for $d_2$, $80\text{ km/h}$ for $v_2$, $34\text{ km}$ for $d_3$  and $100\text{ km/h}$ for $d_3$ in (IV) to find $\Delta t$

    $\begin{array}{c}\Delta t=\left(\frac{16\text{ km}}{90\text{ km/h}}\right)+\left(\frac{8.0\text{ km}}{80\text{ km/h}}\right)+\left(\frac{34\text{ km}}{100\text{ km/h}}\right)\\ =0.177\text{ h}+0.1\text{ h}+0.34\text{ h}\\ =0.617\text{ h}\end{array}$

Substitute $\left(\text{160}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}$ for $\Delta \overrightarrow{v}$ and $0.617\text{ h}$ for $\Delta t$ in (III) to find ${\overrightarrow{a}}_{\text{avg}}$

    $\begin{array}{c}{\overrightarrow{a}}_{\text{avg}}=\frac{\left(\text{160}\text{.7 km/h}\right)\widehat{\text{i}}-\left(7\text{0}\text{.7 km/h}\right)\widehat{\text{j}}}{0.617\text{ h}}\\ =\left(260\widehat{\text{i}}+114.5\widehat{\text{j}}\right){\text{km/h}}^2\end{array}$

Thus, the average acceleration during the entire trip is $\left(260\widehat{\text{i}}+114.5\widehat{\text{j}}\right){\text{km/h}}^2$.