What is the speed of the spaceship as measured by Joe?

[SuperNova1]

Homework Statement

Moe measure the length of his spaceship as 350 m. Joe observes this spaceship as it flies by his position and measures that it took 0.700 s to pass through. What is the speed of the spaceship?

l': the length of the spaceship as measured by MOE = 350m
t: the time as measured by JOE = .7 x10^-6

Homework Equations

i'm not positive on the equations obviously speed = distance/time won't work as joe will not measure the length of the spaceship the same as moe will
i found an equation on the relativity of length and how it changes in relation to the observer outside the moving object which was ... change in t1 = L'/(c-u)
but I'm not too sure whether it's anywhere close to the correct equation

The Attempt at a Solution

i tried distance/time 350/.7*10^-6 = 5*10^8 which can't be right for obvious reasons as the spaceship cannot travel faster than the speed of light
then i tried 0.7*10^-6 = 350/3*10^8 - u
350 = 0.7*10^-6(3*10^8 - u)
350/.7*10-6= 3*10^8 - u
5*10^8 - 3*10^8 = -u
u = -2*10^8
and this answer doesn't look good either
thank you to anyone who can help

 

[HallsofIvy]

You appear to be using length as measured in one reference frame (Moe's) and time as measured in another (Joes's). You can't use those together.

What you can do, I think, is this- let v be Moe's speed relative to Joe. Then the length as measured in Joe's reference frame is 350/\sqrt{1- v^2/c^2}. Now, you can say that "speed" is "distance divided y time"- v= 350/.7\sqrt{1- v^2/c^2}= 500/\sqrt{1- v^2} so that v\sqrt{1- v^/c^2}= 350. Solve that for v.

 

[rude man]

You appear to be using length as measured in one reference frame (Moe's) and time as measured in another (Joes's). You can't use those together.

What you can do, I think, is this- let v be Moe's speed relative to Joe. Then the length as measured in Joe's reference frame is 350/\sqrt{1- v^2/c^2}. Now, you can say that "speed" is "distance divided y time"- v= 350/.7\sqrt{1- v^2/c^2}= 500/\sqrt{1- v^2} so that v\sqrt{1- v^/c^2}= 350. Solve that for v.

The spaceship from Joe's view contracts. You have it expanding. (Lorentz contraction).
Just multiply 350m by √(1 - v²/c²) instead of dividing.

So the idea is: compute the length of the spaceship as measured by Joe, divide by Joe's time, and that will give the (contracted) length as seen by Joe.

 

[SuperNova1]

Thank you for all your help guys i did it again and got a value of 0.988c or 2.994*10^8 which sounds right to me as the lecturer told us it was close to the speed of light, thank you again