Basic Acceleration and Distance Problem

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An electron accelerates from 14,200 m/s to 3.73 x 10^6 m/s over a distance of 1.93 cm, prompting a discussion on how to calculate the time and acceleration. Participants suggest using the kinematic equation v^2 = v_0^2 + 2aΔx to find acceleration and time. The initial and final velocities are clarified, and the distance is converted to meters for accurate calculations. One user calculates an acceleration of approximately 360,432,600,000,000 m/s² but struggles with the time calculation, yielding an incorrect result of 1.0345 x 10^-8 seconds. The conversation emphasizes the importance of correctly identifying variables and applying the right equations in kinematics.
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1. An electron in a crt accelerates from 14200 m/s to 3.73x10^6 m/s over 1.93 cm. how long does the electron take to travel this distance? and What is it's acceleration (in m/s^2)

The supposedly "easiest" question on my physics worksheet has me stumped. I know that I have to convert so it would be 1420000 cm/s and 373000000 cm/s , but I'm am stuck because I do not know the time or acceleration? I just don't know what equation to use?! I appreciate anyones help.

Thank You.
 
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Hi balllla, Welcome to PF.
From the set of kinematic equations, find out the equation, which relates initial velocity, final velocity, displacement and acceleration.
 
LaTeX Code: v^2 = v_0^2 + 2 a \\Delta x

Would I use this equation which would find out the acceleration to later find out the displacement.?
 
I would convert the 1.93 cm to 0.0193 m instead.
 
ok but what actually am I trying to find. Am I trying to find time and are 14200m/s and 3.73 velocity?
 
would this work

3.73x10^6 ^2= 14200m/s^2 + 2a(.0193m)

? what would that give me, my acceleration? but I still don't know how to do the first part of the problem.

many thanks.
 
balllla said:
LaTeX Code: v^2 = v_0^2 + 2 a \\Delta x

Would I use this equation which would find out the acceleration to later find out the displacement.?

I believe you meant "time" instead of "displacement".
Put the velocities into m/s instead of cm/s and put the distance into meters.

Okay. Tell us which of those given numbers you think are the Vi, Vf, X, a, and t. Just label and list them. Then finding an equation should be simple.

You are trying to find the acceleration and the time.
 
vi=14200 m/s
vf=3.73x10^6 m/s
x=.0193m
t=?
a=?
 
balllla said:
would this work

3.73x10^6 ^2= 14200m/s^2 + 2a(.0193m)

? what would that give me, my acceleration? but I still don't know how to do the first part of the problem.

many thanks.
You are correct. It is a large number. Proceed.
 
  • #10
rl.bhat said:
You are correct. It is a large number. Proceed.

Are you sure? Would the v^2 on the left side of the equation be the final velocity or the average velocity? I'm not trying to confuse anyone, just asking.
 
  • #11
OK. My acceleration is approx. 360432600000000 m/s^2. When doing the algebra I get an answer of .00000001 seconds for time, but my online computer software is saying it is wrong, I know that my acceleration is right but how is my time wrong?!
 
  • #12
I am getting 1.0345^-8 s.
 
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