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 [itex]350/\sqrt{1- v^2/c^2}[/itex]. Now, you can say that "speed" is "distance divided y time"- [itex]v= 350/.7\sqrt{1- v^2/c^2}= 500/\sqrt{1- v^2}[/itex] so that [itex]v\sqrt{1- v^/c^2}= 350[/itex]. 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 [itex]350/\sqrt{1- v^2/c^2}[/itex]. Now, you can say that "speed" is "distance divided y time"- [itex]v= 350/.7\sqrt{1- v^2/c^2}= 500/\sqrt{1- v^2}[/itex] so that [itex]v\sqrt{1- v^/c^2}= 350[/itex]. 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

 

[asteriatic]

I would approach this problem by first considering the concept of relative velocity. This means that the speed of an object can be different depending on the observer's frame of reference. In this case, Moe and Joe have different frames of reference - Moe is on the spaceship and Joe is on the ground.

To determine the speed of the spaceship as measured by Joe, we can use the equation v = d/t, where v is the speed, d is the distance, and t is the time. However, since we are dealing with different frames of reference, we need to use the concept of time dilation and length contraction.

The equation you mentioned, t' = L'/(c-u), is correct. This is the equation for time dilation, where t' is the time measured by an observer in a different frame of reference, L' is the length of the object as measured by that observer, c is the speed of light, and u is the relative velocity between the two frames of reference.

In this case, we know that t = 0.7 x 10^-6 s and L' = 350 m. We also know that c = 3 x 10^8 m/s. So we can rearrange the equation to solve for u:

u = (L'/t') - c
= (350/0.7 x 10^-6) - 3 x 10^8
= 5 x 10^11 - 3 x 10^8
= 4.9997 x 10^11 m/s

This is the relative velocity between Moe and Joe's frames of reference. To find the speed of the spaceship as measured by Joe, we can simply subtract this velocity from the speed of light. So the speed of the spaceship as measured by Joe is approximately 3 x 10^8 m/s.

It's important to note that this is just an approximation and may not be entirely accurate due to the limitations of the given information. However, it gives us a general idea of the spaceship's speed as measured by Joe.