Relation between relativistic kinetic energy and particle velocity/distance travelled

The problem that needs to be solved is:

Given that a particle with mass m, has a relativistic kinetic energy T and an average life span $t_{0}$ in it's own frame of reference, express the magnitude of the average distance S traveled by the particle within it's life span.

My strategy so far has been to solve for the velocity from one of the kinetic energy formuli:

$T = (γ - 1) m c^{2} \Rightarrow (a f t e r r e a r r a n g e m e n t) v = c * \sqrt{1 - \frac{m c^{2}}{T + m c^{2}}}$

With the deffinition of γ being:

$γ = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

And then using the attained velocity, I multipliy it by dialated time to get the distance travelled:

$S = v * t = v * γ * t_{o}$

This has however not been yielding the correct answers. I'm wondering if there is something i've missed?

Help in Swedish is also appreciated, if that is preferred by the helper!

Finally managed to solve it: Using the other definition of kinetic energy:

$T = \sqrt{p^{2} c^{2} + m^{2} c^{4}} - m c^{2} w h e r e p = γ * m * v$

Solving for velocity here yields:

$v = \frac{\sqrt{T^{2} + 2 T m c^{2}}}{m c γ}$

Which translates to a distance (same way as before)

$S = \frac{\sqrt{T^{2} + 2 T m c^{2}}}{m c^{2}} * c * t_{0}$

Where I factored out c to get $m c^{2}$ back.

This formula is tested and works!

Svara Avbryt