Time dilation

Time dilation defines, $\Delta t' = \frac{\Delta t}{\sqrt{1-\left ( \frac{v^{2}}{c^{2}} \right )}}$ where $\Delta t$ is the time of particle's reference frame and $\Delta t'$ is the time of our reference frame. Switching $\Delta t=1.00 \times 10^{-7}$ , we obtain

$\Delta t' = \frac{1.00 \times 10^{-7}}{\sqrt{1-\left ( \frac{(0.99c)^{2}}{c^{2}} \right )}}$ $\Rightarrow \Delta t' = \frac{1.00 \times 10^{-7}}{\sqrt{1-0.99^{2}}}$ $\quad \quad \quad \quad = 7.0881 \times 10^{-7}$

So according to our reference frame the particle has traveled a distance $s = 0.99c \times \Delta t' = (0.99 \times 3 \times 10^{8})ms^{-1} \times (7.0881 \times 10^{-7})s$ $=\boxed{210.51657}$

From time dilation, t= (1 x 10^-7)/(1-0.99^2)^1/2......... t=7.0888 x 10^-7............ s=v x t.............. s= 0.99 x 3 x 10^8 x 7.08881 x 10^-7......... s=210............

whereas for the particle frame their life time is unchanged and .. the length of the path got contracted . Lorentz factor plays in both time dilation and length contraction.