91Ó°ÊÓ

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 $v_{BP}=\frac{1}{3}c$.

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.

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

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.

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

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.

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

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: