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## Homework Statement

A spaceship sets off from Earth for a distant destination, travelling in a straight line at uniform speed [tex]\frac{3c}{5}[/tex]. Ten years later a second ship sets off in the same direction at [tex]\frac{4c}{5}[/tex].

i) Consider the distance travelled by each spaceship as a function of time to work out when (in the Earth frame) the second spaceship will have travelled as far as the first.

ii) Calculate which of the ship's captains has aged the most during their journey, and by how much.

## Homework Equations

The lorentz transformations, and [tex]speed = \frac{distance}{time}[/tex]

## The Attempt at a Solution

I have a solution, but it's one of the first SR problems I've done so I am unsure whether it's correct.

Using c = 1 lightyear per year.

i) In earth frame:

distance travelled by first rocket = distance travelled by second rocket

[tex]\frac{3ct}{5} = \frac{4c(t-10)}{5}[/tex]

[tex]3t = 4t - 40[/tex]

[tex]t = 40[/tex]

So the rockets have travelled the same distance after 40 years in the Earth's frame.

ii) Where x is distance travelled in Earth's frame when rockets meet:

x = speed of rocket 1 * time

[tex] = \frac{3}{5} * 40[/tex]

= 24 light years

time elapsed in rocket 1's frame = [tex]\frac{1}{\sqrt{1 - \frac{3^2}{5^2}}} * (40 - \frac{3*24}{5})[/tex]

[tex]=\frac{5}{4} * 25.6[/tex]

= 32 years

time elapsed in rocket 2's frame = [tex]10 + \frac{1}{\sqrt{1 - \frac{4^2}{5^2}}} * (30 - \frac{4*24}{5})[/tex]

[tex]= 10 + \frac{5}{3} * 10.8[/tex]

= 28 years

So the captain of rocket 1 has aged by 4 years more.

Is this the correct way to approach this problem, and is the answer correct?

Thanks for your help : )

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