(a) Use the distance and velocity data to find the rate of expansion as a function of distance.

(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.

Velocity is the rate of change in position of an item in motion as seen from a specific frame of reference and measured by a specific time standard.

The velocity of galaxy 1 relative to galaxy 2 will be equal to the sum of the velocity of galaxy 1 relative to 3 and the velocity of galaxy 3 relative to 2. Galaxy 1 is moving away from galaxy 3 at a certain speed.

The velocity of the galaxy 1 relative to 3 is V₁₃ =$-4500$m/s.

The velocity of the galaxy 3 relative to 2 is V₃₂=$-2200$m/s.

Hence V₂₃ = $2200$ m/s.

Similarly, for galaxy 4:

$\begin{array}{rcl}{R}_4& =& \frac{V}{d}\\ & =& \frac{2830{\displaystyle \raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{190MIY}\\ & =& 14.9\frac{km}{s\times MIY}\end{array}$

Similarly, for galaxy 1:

$\begin{array}{rcl}{R}_1& =& \frac{V}{d}\\ & =& \frac{-4500{\displaystyle \raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{300MIY}\\ & =& -15\frac{km}{s\times MIY}\end{array}$

Similarly, for galaxy 5:

$\begin{array}{rcl}{R}_5& =& \frac{V}{d}\\ & =& \frac{6700{\displaystyle \raisebox{1ex}{$km$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{450MIY}\\ & =& 14.9\frac{km}{s\times MIY}\end{array}$

Taking the average without considering the negative sign

$\begin{array}{rcl}Averagerateofexplanation& =& \frac{14.7+14.9+15+14.9}{4}\\ & =& 14.9\frac{km}{s\times MIY}\end{array}$

The rate of the expansion of the galaxy is $14.9$km/s per million light-years.

The light travels in one year is$=3\times {10}^8\times 60\times 60\times 24\times 365.25$.

$1\mathrm{light}\mathrm{year}=9.47\times {10}^{15}\mathrm{m}$

Hence 1 million light year.

$$\begin{gathered} d=9.47\times {10}^{15}\times {10}^8 \\ d=9.47\times {10}^{21}m \end{gathered}$$

Distance in kilometer

$$\begin{gathered} \mathrm{d}=\frac{9.47\times {10}^{21}}{1000}\mathrm{km} \\ \mathrm{d}=9.47\times {10}^{18}\mathrm{m} \end{gathered}$$

The time taken will be calculated using the equation

$\begin{array}{rcl}V& =& \frac{d}{t}\\ 14.9& =& \frac{9.47\times {10}^{18}}{t}\\ t& =& 6.35\times {10}^{17}s\end{array}$

Time in years can be calculated as;

$\begin{array}{rcl}t& =& \frac{6.35\times {10}^{17}s}{60\times 60\times 24\times 365.25}\\ t& =& 2.01\times {10}^{10}years\end{array}$

Hence the age of the universe is 20 billion years.