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

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Homework Help Overview

The discussion revolves around a physics problem involving the calculation of velocity for a spaceship traveling a distance of six light-years in a time frame of 2.5 years, as measured by the spaceship's clocks. The problem raises questions about the appropriate units for velocity derived from light years.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore how to derive velocity from the given distance and time, questioning the resulting units of light years per year. There are inquiries about the calculations leading to the velocity value of 2.4 and the implications of using the speed of light in the context of the problem.

Discussion Status

Participants are actively engaging with the problem, with some providing alternative reasoning methods and others clarifying their calculations. There is a recognition of the need to understand the relationship between the units of distance and time in the context of the problem.

Contextual Notes

There are discussions about the assumptions made regarding the speed of light and how it relates to the calculations. Some participants mention setting the speed of light to 1 for simplification, indicating a potential approach to unit conversion.

chloeid
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Homework Statement


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."

Homework Equations



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

lelYy.jpg


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!
 
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How did you get 2.4 from that equation?
 
PeroK said:
How did you get 2.4 from that equation?

As t' is 2.5 years and x is 6 light years I substituted them into the equation :)
 
chloeid said:
As t' is 2.5 years and x is 6 light years I substituted them into the equation :)
What about ##c##?
 
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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?
 
kuruman said:
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?
If I do this then I get a speed bigger than C!

PeroK said:
What about ##c##?

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

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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