Question

Another approach to velocity transformations. In Fig. 37-31, reference frames B and C move past reference frame A in the common direction of their $\mathit{x}$axes. Represent the $\mathit{x}$components of the velocities of one frame relative to another with a two-letter subscript. For example, ${\mathbf{\text{v}}}_{\mathbf{AB}}$is the $\mathbf{x}$component of the velocity of A relative to B. Similarly, represent the corresponding speed parameters with two-letter subscripts. For example, ${\mathbf{\beta }}_{\mathbf{AB}}\mathbf{(}{\mathbf{\text{=v}}}_{\mathbf{AB}}\mathbf{/}\mathbf{c}\mathbf{)}$is the speed parameter corresponding to ${\mathbf{\text{v}}}_{\mathbf{AB}}$.

(a) Show that${\mathbf{\beta }}_{\mathbf{AC}}\mathbf{=}\frac{{\mathbf{\beta }}_{\mathbf{AB}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{BC}}}{\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{AB}}{\mathbf{\beta }}_{\mathbf{BC}}}$

Let ${\mathbf{M}}_{\mathbf{AB}}$represent the ratio$\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{\beta }}_{\mathbf{AB}}\mathbf{)}\mathbf{/}\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{\beta }}_{\mathbf{AB}}\mathbf{)}$ , and let${\mathbf{M}}_{\mathbf{BC}}$ and${\mathbf{M}}_{\mathbf{AC}}$ represent similar ratios.

(b) Show that the relation

${\mathbf{M}}_{\mathbf{AC}}\mathbf{=}{\mathbf{M}}_{\mathbf{AB}}{\mathbf{M}}_{\mathbf{BC}}$

is true by deriving the equation of part (a) from it.

Short answer

(a)${\mathrm{\beta }}_{\mathrm{AC}}=\frac{{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}}{(1+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}})}$ is proved.

(b) ${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}{\mathrm{M}}_{\mathrm{BC}}$, is true.

Step-by-step solution

Identification of given data

The given data can be listed below as:

• The velocity of A relative to B is ${\mathrm{v}}_{\mathrm{AB}}$.
• The value of${\mathrm{\beta }}_{\mathrm{AB}}$ is ${\mathrm{\beta }}_{\mathrm{AB}}={\mathrm{v}}_{\mathrm{AB}}/\mathrm{c}$.
• The value of ${\mathrm{M}}_{\mathrm{AB}}$ is ${\mathrm{M}}_{\mathrm{AB}}=(1-{\mathrm{\beta }}_{\mathrm{AB}})/(1+{\mathrm{\beta }}_{\mathrm{AB}})$.
• The value of ${\mathrm{M}}_{\mathrm{BC}}$is ${\mathrm{M}}_{\mathrm{BC}}=(1-{\mathrm{\beta }}_{\mathrm{BC}})/(1+{\mathrm{\beta }}_{\mathrm{BC}})$.
• The value of${\mathrm{M}}_{\mathrm{AC}}$ is ${\mathrm{M}}_{\mathrm{AC}}=(1-{\mathrm{\beta }}_{\mathrm{AC}})/(1+{\mathrm{\beta }}_{\mathrm{AC}})$.

Significance of the velocity

The velocity is described as the distance moved by an object in a particular time. The velocity is directly proportional to the acceleration of an object.

Determination of βAC=βAB+βBC1+βABβBC

(a)

The relation given in the question is expressed as:

${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}{\mathrm{M}}_{\mathrm{BC}}$

Substitute the values in the above equation.

$\begin{array}{c}\frac{(1-{\mathrm{\beta }}_{\mathrm{AC}})}{(1+{\mathrm{\beta }}_{\mathrm{AC}})}=\frac{(1-{\mathrm{\beta }}_{\mathrm{AB}})}{(1+{\mathrm{\beta }}_{\mathrm{AB}})}\frac{(1-{\mathrm{\beta }}_{\mathrm{BC}})}{(1+{\mathrm{\beta }}_{\mathrm{BC}})}\\ (1-{\mathrm{\beta }}_{\mathrm{AC}})(1+{\mathrm{\beta }}_{\mathrm{AB}})(1+{\mathrm{\beta }}_{\mathrm{BC}})=(1+{\mathrm{\beta }}_{\mathrm{AC}})(1-{\mathrm{\beta }}_{\mathrm{AB}})(1-{\mathrm{\beta }}_{\mathrm{BC}})\\ 1-{\mathrm{\beta }}_{\mathrm{AC}}+{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}-{\mathrm{\beta }}_{\mathrm{AC}}{\mathrm{\beta }}_{\mathrm{AB}}-{\mathrm{\beta }}_{\mathrm{AC}}{\mathrm{\beta }}_{\mathrm{BC}}+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}-{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}{\mathrm{\beta }}_{\mathrm{AC}}=1+{\mathrm{\beta }}_{\mathrm{AC}}-{\mathrm{\beta }}_{\mathrm{AB}}-{\mathrm{\beta }}_{\mathrm{BC}}-{\mathrm{\beta }}_{\mathrm{AC}}{\mathrm{\beta }}_{\mathrm{AB}}\\ -{\mathrm{\beta }}_{\mathrm{AC}}{\mathrm{\beta }}_{\mathrm{BC}}+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}{\mathrm{\beta }}_{\mathrm{AC}}\\ -{\mathrm{\beta }}_{\mathrm{AC}}+{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}-{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}{\mathrm{\beta }}_{\mathrm{AC}}={\mathrm{\beta }}_{\mathrm{AC}}-{\mathrm{\beta }}_{\mathrm{AB}}-{\mathrm{\beta }}_{\mathrm{BC}}+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}{\mathrm{\beta }}_{\mathrm{AC}}\end{array}$

Hence, further as:

$\begin{array}{c}2{\mathrm{\beta }}_{\mathrm{AB}}+2{\mathrm{\beta }}_{\mathrm{BC}}=2{\mathrm{\beta }}_{\mathrm{AC}}+2{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}}{\mathrm{\beta }}_{\mathrm{AC}}\\ 2{\mathrm{\beta }}_{\mathrm{AC}}(1+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}})=2{\mathrm{\beta }}_{\mathrm{AB}}+2{\mathrm{\beta }}_{\mathrm{BC}}\\ {\mathrm{\beta }}_{\mathrm{AC}}=\frac{{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}}{(1+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}})}\end{array}$

Thus,${\mathrm{\beta }}_{\mathrm{AC}}=\frac{{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}}{(1+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}})}$ is proved.

Determination of the relation MAC=MABMBC

(b)

As the equation${\mathrm{\beta }}_{\mathrm{AC}}=\frac{{\mathrm{\beta }}_{\mathrm{AB}}+{\mathrm{\beta }}_{\mathrm{BC}}}{(1+{\mathrm{\beta }}_{\mathrm{AB}}{\mathrm{\beta }}_{\mathrm{BC}})}$ has been derived with the help of the above equation ${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}{\mathrm{M}}_{\mathrm{BC}}$, then it can be identified that this equation${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}{\mathrm{M}}_{\mathrm{BC}}$ holds true.

Thus,${\mathrm{M}}_{\mathrm{AC}}={\mathrm{M}}_{\mathrm{AB}}{\mathrm{M}}_{\mathrm{BC}}$ is true.