# How do I work out units of velocity from light years?

1. May 11, 2017

### chloeid

1. The problem statement, all variables and given/known data
I'm currently working through the following question:

"A spaceship travels from Earth to the vicinity of the star that is measured by astronomers on Earth to be six light-years away. The spaceship and its occupants have a total rest mass of 32 000 kg. Assume that the spaceship travels at constant velocity. The time taken as measured by clocks on the spaceship is 2.5 years. Compute the velocity of the spaceship."

2. Relevant equations

I have derived this equation which gives a velocity of 2.4 but I am confused about the units of this.

3. The attempt at a solution

I thought automatically that the units would be light years per year but that doesn't make any sense? Or do the units cancel out somehow?

Thank you!

2. May 11, 2017

### PeroK

How did you get 2.4 from that equation?

3. May 11, 2017

### chloeid

As t' is 2.5 years and x is 6 light years I substituted them into the equation :)

4. May 11, 2017

### PeroK

What about $c$?

5. May 11, 2017

### kuruman

Another way to do this is to reason it out by comparing it with something familiar. A car that takes 3 times as long as another car to go from A to B must be traveling by what fraction of the faster car's speed?

6. May 11, 2017

### chloeid

If I do this then I get a speed bigger than C!

Ah thank you so much! I set c to 1 and got the right answer :)

7. May 11, 2017

### PeroK

If you go back to your original equation and divide by $c$ you get an expression for $\frac{u}{c}$.

The key term on the right hand side is then:

$\frac{c^2t^2}{x^2}$

Now, if $x$ is measured in light years or light seconds, (and $t$ in years or seconds, as appropriate) then you can write this as $c$.years or $c$.seconds and the $c^2$ cancels out.

Alternatively, of course, you can imagine you are using $c=1$ units

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