# Working out distance for a package traveling space

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?
:)

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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

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.

You've mixed up multiplication and division. The package is traveling at $$u=\tfrac{10}{11}c$$

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.

$$d=ut$$

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.

You've mixed up multiplication and division. The package is traveling at $$u=\tfrac{10}{11}c$$

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.

$$d=ut$$

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.
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

That's pretty close, that's the distance the package travels in one day.

That's pretty close, that's the distance the package travels in one day.
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?

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,

$$\frac{10}{11}c = \frac{10}{11}\frac{Lightdays}{day}$$

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

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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,

$$\frac{10}{11}c = \frac{10}{11}\frac{Lightdays}{day}$$

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