Question

You probably did Prob. 12.4 from the point of view of an observer on the ground. Now do it from the point of view of the police car, the outlaws, and the bullet. That is, fill in the gaps in the following table:

Speed of $\to$Relative to $\mathbf{\downarrow }$	Ground	Police	Outlaws	Bullet	Do they escape?
Ground	0	role="math" localid="1654061605668" $\frac{\mathbf{1}}{\mathbf{2}}\mathit{c}$	$\frac{\mathbf{3}}{\mathbf{4}}\mathit{c}$
Police				$\frac{\mathbf{1}}{\mathbf{3}}\mathit{c}$
Outlaws
Bullet

Short answer

The fill in gaps in the following table are filled as follows:

Speed of $\to$Relative to$\downarrow$	Ground	Police	Outlaws	Bullet	Do they escape?
Ground	0	$\frac{1}{2}c$	$\frac{3}{4}c$	$\frac{5}{7}c$	Yes
Police	$-\frac{1}{2}c$	0	$\frac{2}{5}c$	$\frac{1}{3}c$	Yes
Outlaws	$-\frac{3}{4}c$	$-\frac{2}{5}c$	0	$-\frac{1}{13}c$	Yes
Bullet	$-\frac{5}{7}c$	$-\frac{1}{3}c$	$\frac{1}{13}c$	0	Yes

Step-by-step solution

Given information:

Given data:

The velocity of police relative to the ground is $v_{PE}=\frac{1}{2}c$.

The velocity of the outlaws relative to the ground is $v_{OE}=\frac{3}{4}c$.

The velocity of the bullet relative to the police is $v_{BP}=\frac{1}{3}c$.

Determine the velocity of the bullet relative to the ground:

Write the expression to find the velocity of the bullet relative to the ground.

${\mathit{v}}_{\mathbf{B}\mathbf{E}}\mathbf{=}\frac{{\mathbf{v}}_{\mathbf{B}\mathbf{P}}\mathbf{+}{\mathbf{v}}_{\mathbf{P}\mathbf{E}}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{{\mathbf{v}}_{\mathbf{B}\mathbf{P}}{\mathbf{v}}_{\mathbf{P}\mathbf{E}}}{{\mathbf{c}}^{\mathbf{2}}}}}$

Substitute the value of$v_{BP}$ and$v_{PE}$ in the expression.

$$\begin{gathered} v_{BE}=\frac{\left({\displaystyle \frac{1}{3}}c\right)+\left({\displaystyle \frac{1}{2}}c\right)}{1+{\displaystyle \frac{\left({\displaystyle \frac{1}{3}}c\right)\left({\displaystyle \frac{1}{2}}c\right)}{c^2}}} \\ {v}_{BE}=\frac{\left({\displaystyle \frac{5}{6}}c\right)}{\left({\displaystyle \frac{7}{6}}\right)} \\ {v}_{BE}=\frac{5}{6}c\times \frac{6}{7} \\ {v}_{BE}=\frac{5}{7}c \end{gathered}$$

Determine the velocity of outlaws relative to police:

Write the expression to find the velocity of outlaws relative to the police.

$v_{OP}=\frac{v_{OE}-v_{PE}}{1+{\displaystyle \frac{v_{OE}{v}_{PE}}{c^2}}}$

Substitute the value of$v_{OE}$ and$v_{PE}$ in the above expression.

$$\begin{gathered} v_{OP}=\frac{\left({\displaystyle \frac{3}{4}}c\right)-\left({\displaystyle \frac{1}{2}}c\right)}{1+{\displaystyle \frac{\left({\displaystyle \frac{3}{4}}c\right)\left({\displaystyle \frac{1}{2}}c\right)}{c^2}}} \\ {v}_{OP}=\frac{\left({\displaystyle \frac{1}{4}}c\right)}{\left({\displaystyle \frac{5}{8}}\right)} \\ {v}_{OP}=\left(\frac{1}{4}c\right)\times \left(\frac{8}{5}\right) \\ {v}_{OP}=\frac{2}{5}c \end{gathered}$$

Determine the velocity of outlaws relative to a bullet:

Write the expression to find the velocity of outlaws relative to the bullet.

$v_{OB}=\frac{v_{OE}-v_{EB}}{1+{\displaystyle \frac{v_{OE}{v}_{EB}}{c^2}}}$

Substitute the value of$v_{OE}$and$v_{EB}$in the above expression.

$$\begin{gathered} v_{OB}=\frac{\left({\displaystyle \frac{3}{4}}c\right)-\left({\displaystyle \frac{5}{7}}c\right)}{1+{\displaystyle \frac{\left({\displaystyle \frac{3}{4}}c\right)\left({\displaystyle \frac{5}{7}}c\right)}{c^2}}} \\ {v}_{OB}=\frac{\left({\displaystyle \frac{1}{28}}c\right)}{\left({\displaystyle \frac{13}{28}}\right)} \\ {v}_{OB}=\left(\frac{1}{28}c\right)\times \left(\frac{28}{13}\right) \\ {v}_{OB}=\frac{1}{13}c \end{gathered}$$

The velocity of outlaws in each case is greater than the velocity of a bullet, which means they escape.

The fill in gaps in the following table are filled as follows:

Speed of $\to$Relative to $\mathbf{\downarrow }$	Ground	Police	Outlaws	Bullet	Do they escape?
Ground	0	$\frac{1}{2}c$	$\frac{3}{4}c$	$\frac{5}{7}c$	Yes
Police	$-\frac{1}{2}c$	0	$\frac{2}{5}c$	$\frac{1}{3}c$	Yes
Outlaws	$-\frac{3}{4}c$	$-\frac{2}{5}c$	0	$-\frac{1}{13}c$	Yes
Bullet	$-\frac{5}{7}c$	$-\frac{1}{3}c$	$\frac{1}{13}c$	0	Yes