Muon Speed: Stopping Distance & Accel Calculation

[mb85]

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 formula... 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 + volt + 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 that's and incorrect answer.. Can someone check over my work and help me out.

Also, if u don't 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 I am getting is not correct. any help would help...

thanks in advance.

 

[qtp]

one thing that i think is wrong is that the particle is decelerating so you should use -a instead of a which will change your position function and probably give u the right answer

 

[mb85]

hey i figured out the second question i had about the g magnitude.
i didnt realize i divided A/G to get my final answer.

 

[Integral]

I would mark it down for having the incorrect number of significant digits.

Beyond that I did not get the same distance traveled as you did. Your methods are correct, but you appear to have an arithmetic error some where.

I would be bothered by a negative time. Generally a deceleration is considered as negative, but that does not change the magnitude of the result, only the sign. Your magnitude is not correct.

 

[mb85]

hmm i did notice a mistake where i intially put in the velocity for the acceleration in the second part. but i re-worked it and got a new answer.

(0 - 4.56 × 10^6 m/s)/( - 3.48 × 10^14 m/s2)
t = 1.31034483 x 10^-8 s


X = Xo + volt + 1/2(a)(t)^2
so = 0 + (4.56 × 10^6 m/s)(1.31034483 x 10^-8) + 1/2 (- 3.48 × 10^14 m/s2)(1.31034483 x 10^-8)^2
and i get = .029875862 m

 

[Integral]

You need to learn to show the correct number of signifant digits in your answer.

For the second problem you must DIVIDE your computed a by g to find the number of g's

 

[mb85]

sorry. the only reason i am not using the correct sig. figures is because we use an Egrade program which doesn't require it. so my teacher said it would be better to not use them when we use Egrade, otherwise we must.

thanks.

 

[Integral]

Someone needs to slap your teacher along side his head. You need to develop the habit of using the correct number of significant digits. Spewing the entire contents of your calculators display is poor form.