Magnitude of Acceleration

  • Thread starter delfam
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  • #1
delfam
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Homework Statement


electron starting from rest and moving with constant acceleration travels 4.7cm in 8.2ms. What is the magnitude of this acceleration? Answer in km/s^2


Homework Equations


X=Xi + (Vi)t + 1/2at^2


The Attempt at a Solution


Vi = 0 so the equation becomes X = Xi + 1/2at^2
t = 8.2ms and X = 4.7 so 4.7 = Xi + 1/2a(8.2)^2

I feel like I did everything right up to this point but all I need is acceleration to plug in but I'm not sure how to get that.
 
Last edited:

Answers and Replies

  • #2
G01
Homework Helper
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You do not need to plug in acceleration. You are trying to solve for a.

Thus, what you need to plug in is xi, the starting position of the particle.
 
  • #3
delfam
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4.7cm = .000007km and 8.2ms = .0082 seconds

electron is from rest so Vi=0 so the equation is 2(X-Xi)/t^2, I have time and I have X which equals .000047km. How to I find Xi, or is that also zero?
 
Last edited:
  • #4
housemartin
87
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Hello!
problem statement says that: "starting from rest", which means that initial velocity is zero. For initial position Xi you can chose whichever value you like. What is given - the distance moved, that is X-Xi. You can chose X = 4.7 cm only if your Xi = 0.
 
  • #5
delfam
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thanks guys, yeah I got it, it was 1.398 km/s^2
 
  • #6
housemartin
87
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your welcome
 

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