If someone could help solve this I would greatly appreciate it. I am stumped on

Question

If someone could help solve this I would greatly appreciate it. I am stumped on the last part. Here is what I got so far for the problem:

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54x107 light-years from Earth. If the lifetime of a human is taken to be 90 years, a spaceship would need to achieve some minimum speed Vmin to deliver living human being to this galaxy. How close to the speed of light would this miniumum speed be?

Express your answer as the difference between Vmin and the Speed of Light c.

c- Vmin = ______________ m/s

Here is the work I've done so far:

Distance of Galaxy= 2.54x107 light - years

= (2.54x107)(9.46x1015 meters)

=2.40284x1023 m

Let the speed with which a human being shall go there, Vmin=V

time allowed (t0)= 90 years

=2.8386x109 s

Applying the relativity priniciple:

=> (8.988x1016)-v2= (1.181x10-14)(8.988x1016)V

=> (8.988x1016)-v2=1061.4828V

=> v2+1061.4828v-8.988x1016=0

* Use Quadratic Formula to get

v= 2.99798x108 m/s or v= -2.99798x108 m/s

This is the hard part I am having troubles with. Putting it all together

Explanation / Answer

d = 2.54x10^7 light year

1light year = 9.46x10^15 m

d =(2.54x10^7 )( 9.46x10^15)

d = 2.40484x1023 m

v =d/to

t = d/v = 2.40484x1023 /v

tmie to = 90 years = 90*365*24*60*60 s

t0 =2.8386x109 s

In relativistic motion

t =to/[1-(v2/c2)]1/2

[1-(v2/c2)]1/2 = to/t

[1-(v2/c2)] =(to/t)2 =[ (v x2.8386x109 ) /2.40484x1023]2

c2 - v2 = (1.81x10-14)xc2v

Speed of light c =3x10^ 8 m/s

(3x108)2 - v2 = (1.81x10-14)x(3x108)2 xv

9x1016 - v2 = 1061.4828v

v2 +1061.4828v - 9x1016 =0

By solving above equation we get

vmin =2.99798x108m/s

c -vmin =(3x108) -(2.99798x108)

c -vmin = 2.02x105 m/s

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