How Fast Does a Ship Travel to Reach the Nearest Star in 4.25 Years?

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tomeatworld
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Homework Statement


According to observations on Earth, the distance to nearest star is 4.5 light-years. A ship which leaves Earth takes 4.25 years (according to onboard clock) 4.25 years to reach this star. Calculate the speed at which this ship travels.


Homework Equations


(I think)
x' = [tex]\gamma[/tex](x-ut)
t' = [tex]\gamma[/tex](t-ux/c2)
L = L0/[tex]\gamma[/tex]

The Attempt at a Solution


so we know:
x = 4.25 light years, and, t' = 4.25 years.
using length contraction, L (or x') = 4.25 / [tex]\gamma[/tex]]

So i tried using this in x = [tex]\gamma[/tex](x' + ut') but only managed to get 0 = [tex]\gamma[/tex]ut'.

I can't find any other way of reducing variables to obtaining new variables and would love a push in the right direction.

Thanks in advance.
 
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If you want to do this question the Lorentz transformation way, you need to have 2 events, one being leaving Earth, and one being arriving at the star.

So event 1: (x,t)=(0,0) (x',t')=(0,0) so that your coordinates are synchronized at the point of departure. This step is sometimes implicit, but you should account for this for completeness.

Event 2: (x,t)=(4.5ly, ?) (x',t')=(0,4.25years)

Your job is then to use the transforms to get the question mark. In the end v=x/t.

A simpler way to answer this IMO, is to just use length contraction. You know that L'=L/gamma. Speed v, then is just v=L'/t'=L/gamma*t'. Gamma has a v in it as well, and you can solve for v.

This method requires you to keep track of what exactly L and L' mean, and what exactly you are trying to solve for. It can be misused...but it's simpler than using the Lorentz transforms.
 
Using the length contraction method, I managed to get:

v = [tex]\frac{1}{\sqrt{2}}[/tex]c

Any chance of confirmation?
 
Let's see:

L=4.5ly t'=4.25 years

[tex]v=\frac{L}{\gamma t'}=\frac{L\sqrt{1-\frac{v^2}{c^2}}}{t'}[/tex]

[tex]v^2=\frac{L^2(1-\frac{v^2}{c^2})}{t'^2}[/tex]

Solving for v, I get: v=.529c which doesn't seem to match yours. Perhaps you can show your work?