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?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Vf = Vi muan physics

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