Working out distance for a package traveling space

[ghostbuster25]

I have been asked to work out the distance a package has traveled in 5 days if it is traveling 0.909090901c the speed of light.

I know its a simple question but there is a twist i don't get. I worked out

0.909090901/c = 30303030.03 m/s (that is the package is traveling 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 traveling 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?
:)

 

[Mindscrape]

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

 

[ghostbuster25]

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 traveled 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.

 

[RoyalCat]

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.

 

[ghostbuster25]

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 I am still unsure about what i am doing. i think I am overcomplicating it in my head

 

[RoyalCat]

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

 

[ghostbuster25]

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

so I've 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?

 

[AtticusFinch]

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.

Just multiply by 5 and you will have your answer.

 

[ghostbuster25]

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.

Just multiply by 5 and you will have your answer.

ahhh! ok makes sense :)
thanks for the help

 

[AtticusFinch]

No problem you can do the same thing for the other question as well.