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Homework Help: Working out distance for a package traveling space

  1. Mar 31, 2010 #1
    I have been asked to work out the distance a package has travelled in 5 days if it is travelling 0.909090901c the speed of light.

    I know its a simple question but there is a twist i dont get. I worked out

    0.909090901/c = 30303030.03 m/s (that is the package is travelling at 30303030.03 m/s )

    I then worked out there were 432000 seconds in 5 days so just multiplied them together to get 1.309090897*1013 metres.

    Which i am happy with. It then asks me to state it in terms of light days....this is where i am stuck.
    Any help?

    It then asks how long a package travelling the same distance but at a velocity of 0.8035714286c takes. I think i am supposed to do t=d/v but this just gives me a stupid number.
    ANy help with this too?
  2. jcsd
  3. Mar 31, 2010 #2
    You may want to recalculate your velocity, but you can also leave it in terms of c if all you need to do is express the distance as lightdays. Remember .909c is .909*c, not division.

    A light year is how much something can travel going the speed of light in one year, so a light day is how much something can travel going the speed of light in one day.

    1 lightday = c*1day
  4. Mar 31, 2010 #3
    I understood what a light day was and worked it out to be 2.592*1013m in a day. I am just having a mental block on how to relate that to my result for the distance travelled by my package.

    The result i got which seemed way off was 3.258181789*1012 seconds.

    To be honest i think i am making a stupid mistake somewhere and its probably staring at me in the face.
  5. Mar 31, 2010 #4
    You've mixed up multiplication and division. The package is traveling at [tex]u=\tfrac{10}{11}c[/tex]

    Dividing by the velocity of light doesn't give you units of velocity, mind you.

    The formula you should keep in mind is that of distance traveled as a function of time, for an object in uniform motion.


    Finding the distance in meters is just a straightforward matter of finding the velocity, u and the time, t, in the appropriate units.

    As for the second part of the question, a light-day, much like a light-year, is defined as the distance that light travels during a day.

    That is to say, that it is the distance an object traveling at velocity c would travel during 1 day.
  6. Mar 31, 2010 #5
    Ok i see where i went wrong with the distance and now have a figure of
    2.356363615*1013 for the distance travelled. I have used the form t=d/u but im still unsure about what i am doing. i think im overcomplicating it in my head
  7. Mar 31, 2010 #6
    That's pretty close, that's the distance the package travels in one day.
  8. Mar 31, 2010 #7
    so ive worked out the distance the package travels in a day and i know how far light travels in a day. do i just divide to get the result?
  9. Mar 31, 2010 #8
    Are we looking for the distance the object travels with respect to an observer on Earth or the package itself?

    Assuming we are going for the Earth observer you know,

    [tex] \frac{10}{11}c = \frac{10}{11}\frac{Lightdays}{day} [/tex]

    This is because c is equal to 1 Lightday/day (light travels a distance of one lightday per day) really convenient.

    Just multiply by 5 and you will have your answer.
    Last edited: Mar 31, 2010
  10. Mar 31, 2010 #9
    ahhh!!! ok makes sense :)
    thanks for the help
  11. Mar 31, 2010 #10
    No problem you can do the same thing for the other question as well.
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