A muon (an elementary particle) enters a region with a speed of 4.56 × 10^6 m/s and then is slowed at the rate of 3.48 × 10^14 m/s2. How far does the muon take to stop? so i use the forumla... Vf = Vi + at so then Vf-Vi/a = t (0 - 4.56 × 10^6 m/s)/3.48 × 10^14 m/s2 t = - 1.31034483 x 10^-8 s so then i use the position formula X = Xo + Vot + 1/2(a)(t)^2 so = 0 + (4.56 × 10^6 m/s)(- 1.31034483 x 10^-8) + 1/2 (4.56 × 10^6 m/s)(- 1.31034483 x 10^-8)^2 and i get = - .0597517239 m But of course... Egrade says thats and incorrect answer.. Can someone check over my work and help me out. Also, if u dont mind. A world's land speed record was set by Colonel John P. Stapp when in March 1954 he rode a rocket-propelled sled that moved along a track at 1020 km/h. He and the sled were brought to a stop in 1.4 s. In g units, what magnitude acceleration did he experience while stopping? So i realize 1g = 9.8m/s^2 so if i use the formula Vf = Vi + at and get a. how do i solve for g? i know you multiply by whatever ur g is and then mulitply that by 9.8. but the answer im getting is not correct. any help would help... thanks in advance.