Relation between s=d/t and length contraction ?

[Frank_Mennella]

Homework Statement

A 16m long vessel moved passed you at a relative speed of 0.86c. What was the length of the vessel recorded by your measuring device?

Homework Equations

Speed = distance/time
t=t_o/√1-v²/c²
L=L_o⋅√1-v²/c²

The Attempt at a Solution

Ok, so I can just sub the 16m and 0.86c in the length contraction formula and get L≈8.16m

However, I wasn't aware of the length contraction formula, but only of the time dilation formula, which I used (with s=d/t) to find my answer:

So,
t= d/s
t= 16/(0.86 x 3 x10⁸)
t=6.2x10^-8
Therefore, we know the time it takes the particle to travel the distance, (reference frame of the particle)
We can use the time dilation formula to find t_o (reference frame of stationary observant)

6.2x10^-8=t_o/√1-(0.86x3x10⁸)²/(3x10⁸)²

t_o=6.2x10^-8⋅√1-(0.86x3x10⁸)²/(3x10⁸)²

t_o=3.16x10^-8 This is the time it takes the particle to travel the distance at the reference frame of a stationary observant.

Therefore, to find the length of the vessel I use:
d= s⋅t
d= 0.86x3x10⁸⋅3.16x10^-8
d=8.15m (Remember that I have rounded up some numbers)

Is that a second way to find the answer? Does it make sense? Me and my professor were looking for an explanation and would like to know why does or doesn't this work.

 

[jbriggs444]

Therefore, we know the time it takes the particle to travel the distance, (reference frame of the particle)

In the reference frame of the particle the particle does not travel at all.

You are throwing a lot of variable names around without clearly defining what any of them denote. That's a sure way to get into trouble with Special Relativity. And you've never once considered the relativity of simultaneity. That's an even more certain way to reason yourself into a paradox.