Stopping a Muon: Solving the Equations

  • Thread starter Thread starter lauriecherie
  • Start date Start date
  • Tags Tags
    Muon
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a muon that enters a region with a specified speed and is subjected to a deceleration. Participants are tasked with determining the distance the muon travels before coming to a stop, utilizing various kinematic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of different kinematic equations to find the distance traveled, questioning the correctness of their calculations and the signs of their values. There is an exploration of whether time is necessary for the solution.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and identifying potential errors in the application of equations. Some guidance has been offered regarding the signs of values and the proper use of equations, but no consensus on the final answer has been reached.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the initial speed and acceleration of the muon. There is an emphasis on ensuring the correct interpretation of the final velocity and the implications of negative acceleration.

lauriecherie
Messages
44
Reaction score
0

Homework Statement



A muon (an elementary particle) enters a region with a speed of 4.20 106 m/s and then is slowed at the rate of 1.10 1014 m/s2.
(a) How far does the muon take to stop?
________ m


Homework Equations



x(t)= initial position + final velocity * time

v(t)= (acceleration * time) + initial velocity

x(t)= .5 * (acceleration * (time^2)) + (initial velocity * time) + inital position

x= initial position * (average velocity * time)

average velocity= (final velocity - initial velocity) / (2)

(final velocity^2) - (initial velocity^2) = 2 * acceleration * change in position




The Attempt at a Solution



I came out with this as my seconds, t, .00000003818181818. Is this correct for the t value and where should I put this in at?
 
Physics news on Phys.org
Your time looks to be about correct. You can substitute it into your equation number three.

Alternatively, you can solve this one without time with another one of the equations you mentioned =)
 
Some of the equations you typed are wrong. Acceleration and initial velocity are given. What is the final velocity of the particle (hint: read the problem)? Which kinematical equation should be used to calculate the distance traveled when acceleration and the velocities are known?
 
By using the last equation I listed, I came out with -1.90909091 x 10^-8. This doesn't seem correct? Is this value correct and should it be a positive number?
 
Umm, no that is not correct. Perhaps you substituted something in wrong?

As far as signs are concerned remember acceleration opposes the velocity, so its sign is negative.
 
Hummm... this is what I did:

0^2 - (4.20 x 10^6)^2 = 2 * (-1.1 x 10^14)* change in X

{4200000/((-1.4 x 10^14) * 2)}

Change in X = -1.90909091 x 10^ -8.

Maybe I entered it into my calculator incorrectly?
 
Looks like you forgot to square the initial velocity. You have the ^2 written in your first step, but you drop it in the 2nd. Also, you dropped the - sign from steps 1 to 2.

Looks like those are your problems,
Cheers!
 
Let's try this again...


-.0801818182?
 
Again, you forgot the sign, but yes.
 
  • #10
+.0801818182 m Thanks so much for your help!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K