Question

Lisa in her Lamborghini accelerates at the rate of$\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{)}\mathbf{ }\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, while Jill in her Jaguar accelerates at$\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{j}}\mathbf{)}\mathbf{}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. They both start from rest at the origin of an$\mathit{x}\mathbf{,}\mathit{y}$coordinate system. After$5.00\text{s}$,

(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?

Short answer

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

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

(c) The relative acceleration of Lisa with respect to Jill is $(2.00i-5.00\hat{j}) \mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

The first law of motion.

The first law of motion is given by

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

Here,${\overrightarrow{\mathbf{u}}}_{\mathbf{L}}$is the initial velocity,$\mathit{t}$is the time,${\overrightarrow{\mathbf{v}}}_{\mathbf{L}}$is the final velocity of and${\overrightarrow{\mathbf{a}}}_{\mathbf{L}}$is the acceleration.

(a) Determine the Lisa’s speed with respect to Jill:

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

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

role="math" localid="1663608503719" ${\overrightarrow{v}}_{\mathrm{L}}={\overrightarrow{u}}_{\mathrm{L}}+{\overrightarrow{a}}_{\mathrm{L}}t$

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

Substitute$0$for${\overrightarrow{u}}_{\mathrm{L}},5.00\sec$for$t$and$(3.00i-2.00\hat{j})\mathrm{m}/{\mathrm{s}}^2$for role="math" localid="1663608209397" ${\overrightarrow{a}}_{\mathrm{L}}$to find${\overrightarrow{v}}_{\mathrm{L}}$.

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

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

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

Here, ${\overrightarrow{u}}_{\mathrm{J}}$is the initial velocity of Jill, ${\overrightarrow{v}}_{\mathrm{J}}$is the final velocity of Jill and ${\overrightarrow{a}}_{\mathrm{J}}$is the acceleration of Jill. Substitute $0$for ${\overrightarrow{u}}_{\mathrm{J}},5.00\sec$for $t$and $(1.00i+3.00\hat{j})\mathrm{m}/{\mathrm{s}}^2$forto find the.

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

The relative velocity of Lisa with respect to Jill is given by

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

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

Substitute $(15.0i-10.0\hat{j})\mathrm{m}/\mathrm{s}$for,$(5.00i+15.0\hat{j})\mathrm{m}/\mathrm{s}$for and find.

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

Find the magnitude of the above found velocity

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

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

(b) Determine the distance between Lisa and Jill:

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

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

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

Substitute $0\mathrm{m}/\mathrm{s}$for ${\overrightarrow{u}}_{\mathrm{L}}5.00\sec$for $t$and localid="1663609022105" $\left(3.00i-2.00\hat{j}\right)\mathrm{m}/{\mathrm{s}}^2$for ${\overrightarrow{a}}_{\mathrm{L}}$to find the .

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

Write the expression for the separation between these persons

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

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

Substitute $(37.5i-25\hat{j})\mathrm{m}$forand $(12.5i-37.5\hat{j})\mathrm{m}$for to find ${\overrightarrow{s}}_{\mathrm{LJ}}$.

$\begin{array}{c}{\overrightarrow{s}}_{\mathrm{LJ}}=(37.5\hat{i}-25\hat{j})\mathrm{m}-(12.\hat{i}-37.5\hat{j})\mathrm{m}\\ =(25\hat{i}-62.5\hat{j})\mathrm{m}\end{array}$

$\begin{array}{c}\left|{\overrightarrow{s}}_{\mathrm{LJ}}\right|=\sqrt{{25}^2+{(-62.5)}^2}\\ =67.3\mathrm{m}\end{array}$

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

(c) Determine the Lisa's acceleration relative to Jill.

Write the formula to calculate Lisa acceleration relative to Jill

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

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

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

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

Therefore, Lisa's acceleration relative to Jill is $(2.00\hat{i}-5.00\hat{j}) \mathrm{m}/{\mathrm{s}}^2$.