Chapter 4, Problem 58PQ

Two bicyclists in a sprint race begin from rest and accelerate away from the origin of an x–y coordinate system. Miguel’s acceleration is given by (−0.700 $\hat{i}$ + 1.00 $\hat{j}$) m/s², and Lance’s acceleration is given by (1.20 $\hat{i}$ + 0.300 $\hat{j}$) m/s². a. What is Miguel’s acceleration with respect to Lance? b. What is Miguel’s speed with respect to Lance after 4.50 s have elapsed? c. What is the distance separating Miguel and Lance after 4.50 s have elapsed?

To determine

Find the Miguel’s acceleration with respect to Lance.

Answer to Problem 58PQ

The Miguel’s acceleration with respect to Lance is $\left(-1.90\widehat{i}+0.700\widehat{j}\right){\text{m/s}}^2$.

Explanation of Solution

Integrating the given acceleration will give the velocity of Miguel and Lance as they both starts from rest.

 ${\overrightarrow{v}}_m=\left(-0.700t\widehat{i}+1.00t\widehat{j}\right)\text{m/s}$

${\overrightarrow{v}}_L=\left(-1.20t\widehat{i}+0.300t\widehat{j}\right)\text{m/s}$

Integrating the velocity will give the position of Miguel and Lance as they both starts from rest.

${\overrightarrow{r}}_m=\left(-0.350t^2\widehat{i}+0.500t^2\widehat{j}\right)\text{m}$

${\overrightarrow{r}}_L=\left(0.600t^2\widehat{i}+0.150t^2\widehat{j}\right)\text{m}$

Write the equation for the acceleration of Miguel relative to Lance.

${\left({\overrightarrow{a}}_M\right)}_L={\left({\overrightarrow{a}}_M\right)}_G+{\left({\overrightarrow{a}}_G\right)}_L={\left({\overrightarrow{a}}_M\right)}_G-{\left({\overrightarrow{a}}_L\right)}_G$ (I)

Here, ${\left({\overrightarrow{a}}_M\right)}_L$ is the acceleration of Miguel with respect to Lance, ${\left({\overrightarrow{a}}_M\right)}_G$ is the acceleration of Miguel with respect to ground, ${\left({\overrightarrow{a}}_G\right)}_L$ is the acceleration of ground with respect to Lance and ${\left({\overrightarrow{a}}_L\right)}_G$ is the acceleration of Lance with respect to ground.

Conclusion:

Substitute $\left(-0.700\widehat{i}+1.00\widehat{j}\right){\text{m/s}}^2$ for ${\left({\overrightarrow{a}}_M\right)}_G$ and $\left(1.20\widehat{i}+0.300\widehat{j}\right){\text{m/s}}^2$ for ${\left({\overrightarrow{a}}_L\right)}_G$ in equation I.

$\begin{array}{c}{\left({\overrightarrow{a}}_M\right)}_L=\left(\left(-0.700\widehat{i}+1.00\widehat{j}\right){\text{m/s}}^2\right)-\left(\left(1.20\widehat{i}+0.300\widehat{j}\right){\text{m/s}}^2\right)\\ =\left(-1.90\widehat{i}+0.700\widehat{j}\right){\text{m/s}}^2\end{array}$

Therefore, the Miguel’s acceleration with respect to Lance is $\left(-1.90\widehat{i}+0.700\widehat{j}\right){\text{m/s}}^2$.

To determine

Find the Miguel’s speed with respect to Lance after $4.50\text{ s}$ elapsed.

Answer to Problem 58PQ

The Miguel’s speed with respect to Lance after $4.50\text{ s}$ elapsed is $\text{9}\text{.11 m/s}$.

Explanation of Solution

Write the equation for Miguel’s velocity with respect to Lance.

${\left({\overrightarrow{v}}_M\right)}_L={\left({\overrightarrow{v}}_M\right)}_G+{\left({\overrightarrow{v}}_G\right)}_L={\left({\overrightarrow{v}}_M\right)}_G-{\left({\overrightarrow{v}}_L\right)}_G$ (II)

Here, ${\left({\overrightarrow{v}}_M\right)}_L$ is the velocity of Miguel with respect to Lance, ${\left({\overrightarrow{v}}_M\right)}_G$ is the velocity of Miguel with respect to ground, ${\left({\overrightarrow{v}}_G\right)}_L$ is the velocity of ground with respect to Lance and ${\left({\overrightarrow{v}}_L\right)}_G$ is the velocity of Lance with respect to ground.

Conclusion:

Substitute $\left(-0.700t\widehat{i}+1.00t\widehat{j}\right)\text{m/s}$ for ${\left({\overrightarrow{v}}_M\right)}_G$ and $\left(-1.20t\widehat{i}+0.300t\widehat{j}\right)\text{m/s}$ for ${\left({\overrightarrow{v}}_L\right)}_G$ in equation II.

$\begin{array}{c}{\left({\overrightarrow{v}}_M\right)}_L=\left(\left(-0.700t\widehat{i}+1.00t\widehat{j}\right)\text{m/s}\right)-\left(\left(-1.20t\widehat{i}+0.300t\widehat{j}\right)\text{m/s}\right)\\ =\left(-1.90t\widehat{i}+0.700t\widehat{j}\right)\text{m/s}\end{array}$

Substitute $4.50\text{ s}$ for $t$ in the above equation.

$\begin{array}{c}{\left({\overrightarrow{v}}_M\right)}_L=\left(-1.90\left(4.50\text{ s}\right)\widehat{i}+0.700\left(4.50\text{ s}\right)\widehat{j}\right)\text{m/s}\\ =\left(-8.55\widehat{i}+3.15\widehat{j}\right)\text{m/s}\end{array}$

Then the magnitude of velocity is,

$\begin{array}{c}v=\sqrt{{\left(-8.55\text{ m/s}\right)}^2+{\left(3.15\text{ m/s}\right)}^2}\\ =9.11\text{ m/s}\end{array}$

Therefore, the Miguel’s speed with respect to Lance after $4.50\text{ s}$ elapsed is $\text{9}\text{.11 m/s}$.

To determine

Find the distance between Miguel and Lance after $4.50\text{ s}$ elapsed.

Answer to Problem 58PQ

The distance between Miguel and Lance after $4.50\text{ s}$ elapsed is $\text{20}\text{.5 m}$.

Explanation of Solution

Write the equation for Miguel’s position with respect to Lance.

${\overrightarrow{r}}_{ML}={\overrightarrow{r}}_M-{\overrightarrow{r}}_L$ (III)

Here, ${\overrightarrow{r}}_{ML}$ is the Miguel’s position with respect to Lance, ${\overrightarrow{r}}_M$ is the Miguel’s position and ${\overrightarrow{r}}_L$ is the Lance position.

Conclusion:

Substitute $\left(-0.350t^2\widehat{i}+0.500t^2\widehat{j}\right)\text{m}$ for ${\overrightarrow{r}}_M$ and $\left(0.600t^2\widehat{i}+0.150t^2\widehat{j}\right)\text{m}$ for ${\overrightarrow{r}}_L$ in equation III.

$\begin{array}{c}{\overrightarrow{r}}_{ML}=\left(\left(-0.350t^2\widehat{i}+0.500t^2\widehat{j}\right)\text{m}\right)-\left(\left(0.600t^2\widehat{i}+0.150t^2\widehat{j}\right)\text{m}\right)\\ =\left(-0.950t^2\widehat{i}+0.350t^2\widehat{j}\right)\text{m}\end{array}$

Substitute $4.50\text{ s}$ for $t$ in the above equation.

$\begin{array}{c}{\overrightarrow{r}}_{ML}=\left(-0.950{\left(4.50\text{ s}\right)}^2\widehat{i}+0.350{\left(4.50\text{ s}\right)}^2\widehat{j}\right)\text{m}\\ =\left(-19.2\widehat{i}+7.09\widehat{j}\right)\text{m}\end{array}$

Then the distance between Miguel and Lance is,

$\begin{array}{c}d=\sqrt{{\left(-19.2\text{ m}\right)}^2+{\left(\text{7}\text{.09 m}\right)}^2}\\ =20.5\text{ m}\end{array}$

Therefore, the distance between Miguel and Lance after $4.50\text{ s}$ elapsed is $\text{20}\text{.5 m}$.