A Lincoln Continental is twice as long as a VW Beetle, when they are at rest. As the Continental overtakes the VW, going through a speed trap, a (stationary) policeman observes that they both have the same length. The VW is going at half the speed of light. How fast is the Lincoln going? (Leave your answer as a multiple of c.)

Let $L_c$and $L_v$be the length of Continental and VW Beetle, respectively.

It is given that the continental length is twice the length of the VW Beetle.

$L_c=2L_v$

Now, consider$L_c^{,}$and $L_v^{,}$be the length of the Continental and VW Beetle observed by the policeman, respectively.

Write the expression for the length of Continental observed by a policeman.

${\mathbf{L}}_{\mathbf{c}}^{\mathbf{,}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{c}}}{{\mathbf{Y}}_{\mathbf{c}}}$ …… (1)

Write the expression for the length of the VW Beetle observed by the policeman.

${\mathbf{L}}_{\mathbf{v}}^{\mathbf{,}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{v}}}{{\mathbf{Y}}_{\mathbf{v}}}$ …… (2)

Equate equations (1) and (2).

localid="1655813970746" $$\begin{gathered} L_c^{,}=L_v^{,} \\ \frac{L_c}{y_c}=\frac{L_v}{y_v} \\ \frac{2L_v}{y_c}=\frac{L_v}{y_v} \end{gathered}$$

$\frac{\mathbf{2}}{{\mathbf{Y}}_{\mathbf{c}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Y}}_{\mathbf{v}}}$ .....(3)

It is given that $v=\frac{C}{2}$.

Let the speed of Continental be$v^{,}$

Write the expression for the relativistic speed of Continental.

$y_c=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{v\text{'}}{C}}\right)}^2}}$

Write the expression for the relativistic speed of the VW Beetle.

$y_v=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{C^2}}}}$

Substitute the value of $y_c$and $y_v$in equation (3).

$$\begin{gathered} \frac{{\displaystyle \frac{2}{1}}}{\begin{array}{c}\sqrt{1-{\left({\displaystyle \frac{v^{\text{'}}}{c}}\right)}^2}\end{array}}=\frac{{\displaystyle \frac{1}{1}}}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ 2\sqrt{1-\left(\frac{v^{\text{'}}}{c}\right)}=\sqrt{1-{\left(\frac{1}{2}\right)}^2} \\ \sqrt{1-\left(\frac{v^{\text{'}}}{c}\right)}=\frac{\sqrt{3}}{4} \end{gathered}$$

Squaring on both sides,

$$\begin{gathered} {\left[\sqrt{1-{\left(\frac{v^{\text{'}}}{c}\right)}^2}\right]}^2={\left(\frac{\sqrt{3}}{4}\right)}^2 \\ 1-{\left(\frac{v^{\text{'}}}{c}\right)}^2=\frac{3}{16} \\ \frac{3}{16}={\left(\frac{v^{\text{'}}}{c}\right)}^2 \\ {v}^{\text{'}}=\frac{\sqrt{13}}{4}c \end{gathered}$$

Therefore, the speed of the Lincoln Continental is $\frac{\sqrt{13}}{4}c$.