Question

A muon (an elementary particle) enters a region with a speed of 5.55 106 m/s and then is slowed at the rate of 1.25 1014 m/s2.
(a) How far does the muon take to stop?

Answer 1

I would suggest using the kinematic equation:

Vf^2 = Vi^2 + 2 A X

Take Vf = 0 m/s, since the muon will come to a stop.
Take A = - 1.25 x 10^14 m/s^2
Take Vi = 5.55 x 10^6 m/s

Solve for X.