Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
