Chapter 4, Problem 4.61AP

Lisa in her Lamborghini accelerates at $(3.00\hat{i}-2.00\hat{j})\text{m}/{\text{s}}^2$, while Jill in her Jaguar accelerates at $(1.00\hat{i}-3.00\hat{j})\text{m}/{\text{s}}^2$. They both start from rest at the origin. After 5.00 s, (a) what is Lisa’s speed with respect to Jill. (b) how far apart are they, and (c) what is Lisa’s acceleration relative to Jill?

To determine

The relative speed of Lisa with respect to Jill after $5.00\sec$.

Answer to Problem 4.61AP

The relative speed of Lisa with respect to Jill after $5.00\sec$ is $26.9\text{m}/\text{s}$.

Explanation of Solution

The acceleration of Lisa in her Lamborghini is $\left(3.00\widehat{\text{i}}-2.00\widehat{\text{j}}\right)\text{m}/{\text{s}}^{\text{2}}$ and acceleration of Jill is $\left(1.00\widehat{\text{i}}+3.00\widehat{\text{j}}\right)\text{m}/{\text{s}}^{\text{2}}$.

Write the expression for the first law of motion to calculate the final velocity of Lisa’s Lamborghini

    ${\overrightarrow{v}}_{\text{L}}={\overrightarrow{u}}_{\text{L}}+{\overrightarrow{a}}_{\text{L}}t$

Here, ${\overrightarrow{u}}_{\text{L}}$ is the initial velocity of, $t$ is the time, ${\overrightarrow{v}}_{\text{L}}$ is the Lisa’s final velocity of and ${\overrightarrow{a}}_{\text{L}}$ is the Lisa’s acceleration.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{u}}_{\text{L}}$, $5.00\sec$ for $t$  and $\left(3.00i-2.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_{\text{L}}$ to find ${\overrightarrow{v}}_{\text{L}}$.

    $\begin{array}{c}{\overrightarrow{v}}_{\text{L}}=0\text{m}/\text{s}+\left(3.00i-2.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}\times 5.00\sec \\ =\left(15.0i-10.0\widehat{j}\right)\text{m}/\text{s}\end{array}$

Write the expression for the first law of motion to calculate the final velocity of Jill’s Jaguar

    ${\overrightarrow{v}}_{\text{J}}={\overrightarrow{u}}_{\text{J}}+{\overrightarrow{a}}_{\text{J}}t$

Here, ${\overrightarrow{u}}_{\text{J}}$ is the initial velocity of Jill, ${\overrightarrow{v}}_{\text{J}}$ is the final velocity of Jill and ${\overrightarrow{a}}_{\text{J}}$ is the acceleration of Jill.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{u}}_{\text{J}}$, $5.00\sec$ for $t$ and $\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_{\text{J}}$ to find the ${\overrightarrow{v}}_{\text{J}}$.

    $\begin{array}{c}{\overrightarrow{v}}_{\text{J}}=0\text{m}/\text{s}+\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}\times 5.00\sec \\ =\left(5.0i+15.0\widehat{j}\right)\text{m}/\text{s}\end{array}$

The relative velocity of Lisa with respect to Jill

    ${\overrightarrow{v}}_{\text{LJ}}={\overrightarrow{v}}_{\text{L}}-{\overrightarrow{v}}_{\text{J}}$

Here, ${\overrightarrow{v}}_{\text{LJ}}$ is the relative velocity of Lisa with respect to Jill.

Conclusion:

Substitute $\left(15.0i-10.0\widehat{j}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_{\text{L}}$ and $\left(5.00i+15.0\widehat{j}\right)\text{m}/\text{s}$ for ${\overrightarrow{v}}_{\text{J}}$ to find ${\overrightarrow{v}}_{\text{LJ}}$.

  $\begin{array}{c}{\overrightarrow{v}}_{\text{LJ}}=\left(15.0i-10.0\widehat{j}\right)\text{m}/\text{s}-\left(5.00i+15.0\widehat{j}\right)\text{m}/\text{s}\\ =\left(10.0i-25.0\widehat{j}\right)\text{m}/\text{s}\end{array}$

Find the magnitude of the above found velocity

  $\begin{array}{c}\left|{\overrightarrow{v}}_{\text{LJ}}\right|=\sqrt{{10.0}^2+{\left(-25.0\right)}^2}\\ =26.9\text{m}/\text{s}\end{array}$

Therefore, the Lisa’s speed with respect to Jill after $5.00\sec$ is $26.9\text{m}/\text{s}$.

To determine

The distance between Lisa and Jill.

Answer to Problem 4.61AP

The distance between Lisa and Jill is $67.3\text{m}$.

Explanation of Solution

Write the formula for second law of motion to calculate the displacement of Lisa’s Lamborghini

    ${\overrightarrow{s}}_{\text{L}}={\overrightarrow{u}}_{\text{L}}t+\frac{1}{2}{\overrightarrow{a}}_{\text{L}}{t}^2$

Here, ${\overrightarrow{s}}_{\text{L}}$ is the position of Lisa’s.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{u}}_{\text{L}}$ $5.00\sec$ for $t$ and $\left(3.00i-2.00\widehat{j}\text{m}/{\text{s}}^{\text{2}}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_{\text{L}}$ to find the ${\overrightarrow{s}}_{\text{L}}$.

    $\begin{array}{c}{\overrightarrow{s}}_{\text{L}}=0\text{m}/\text{s}t+\frac{1}{2}\left(3.00i-2.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}{\left(5.00\sec \right)}^2\\ =\left(37.5i-25\widehat{j}\right)\text{m}\end{array}$

Write the formula for second law of motion to calculate the displacement of Jill’s Jaguar

    ${\overrightarrow{s}}_J={\overrightarrow{u}}_{\text{J}}t+\frac{1}{2}{\overrightarrow{a}}_{\text{J}}{t}^2$

Here, ${\overrightarrow{s}}_{\text{J}}$ is the position of Jill’s Jaguar.

Substitute $0\text{m}/\text{s}$ for ${\overrightarrow{u}}_{\text{J}}$ $5.00\sec$ for $t$ and $\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_{\text{J}}$ to find ${\overrightarrow{s}}_{\text{J}}$.

    $\begin{array}{c}{\overrightarrow{s}}_{\text{J}}=0\text{m}/\text{s}t+\frac{1}{2}\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}{\left(5.00\sec \right)}^2\\ =\left(12.5i-37.5\widehat{j}\right)\text{m}\end{array}$

Write the expression for the separation between these persons

    ${\overrightarrow{s}}_{\text{LJ}}={\overrightarrow{s}}_{\text{L}}-{\overrightarrow{s}}_{\text{J}}$

Here, ${\overrightarrow{s}}_{\text{LJ}}$ is the relative position of Lisa with respect to Jill.

Substitute $\left(37.5i-25\widehat{j}\right)\text{m}$ for ${\overrightarrow{s}}_{\text{L}}$ and $\left(12.5i-37.5\widehat{j}\right)\text{m}$ for ${\overrightarrow{s}}_{\text{J}}$ to find ${\overrightarrow{s}}_{\text{LJ}}$.

    $\begin{array}{c}{\overrightarrow{s}}_{\text{LJ}}=\left(37.5i-25\widehat{j}\right)\text{m}-\left(12.5i-37.5\widehat{j}\right)\text{m}\\ \text{=}\left(25i-62.5\widehat{j}\right)\text{m}\\ \left|{\overrightarrow{s}}_{\text{LJ}}\right|=\sqrt{{25}^2+{\left(-62.5\right)}^2}\\ =67.3\text{m}\end{array}$

Conclusion:

Therefore, the distance between Lisa and Jill is $67.3\text{m}$.

To determine

The relative acceleration of Lisa with respect to Jill.

Answer to Problem 4.61AP

The relative acceleration of Lisa with respect to Jill is $\left(2.00i-5.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Write the formula to calculate Lisa acceleration relative to Jill

    ${\overrightarrow{a}}_{\text{LJ}}={\overrightarrow{a}}_{\text{L}}-{\overrightarrow{a}}_{\text{J}}$

Here, ${\overrightarrow{a}}_{\text{LJ}}$ is the relative acceleration of Lisa with respect to Jill.

Conclusion:

Substitute $\left(3.00i-2.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_L$ and $\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$ for ${\overrightarrow{a}}_J$ to find ${\overrightarrow{a}}_{LJ}$.

    $\begin{array}{c}{\overrightarrow{a}}_{\text{LJ}}=\left(3.00i-2.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}-\left(1.00i+3.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}\\ =\left(2.00i-5.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, Lisa’s acceleration relative to Jill is $\left(2.00i-5.00\widehat{j}\right)\text{m}/{\text{s}}^{\text{2}}$