Answer
0.1 m
Work Step by Step
We know that the particle has an initial velocity ($v_{0}$) of 5 x 10$^6$ m/s, an end velocity ($v$) of 0 m/s, and a rate of deceleration ($a$) of 1.25 x 10$^{14}$ m/s$^2$. Since we only need to calculate the distance travelled ($x$), our starting position ($x_{0}$) can be 0.
We then plug in those values into $v^2=v_{0}^2+2a(x-x_{0})$ and solve for $x$. Therefore,
$x=-\frac{(5\times10^6)^2}{(2)(-1.25\times10^{14})}=0.1$m
