Basic Acceleration and Distance Problem

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Homework Help Overview

The problem involves an electron accelerating in a cathode ray tube (CRT) from an initial velocity of 14200 m/s to a final velocity of 3.73x10^6 m/s over a distance of 1.93 cm. Participants are tasked with finding the time taken for this distance and the acceleration of the electron.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate kinematic equations to use, particularly questioning the relationship between initial velocity, final velocity, displacement, and acceleration.
  • There is a focus on unit conversion, with some participants suggesting converting velocities to m/s and distance to meters.
  • Questions arise about the definitions of variables and the correct interpretation of the kinematic equation.
  • Some participants express confusion about the values they are trying to find, specifically distinguishing between time and acceleration.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify their understanding of the variables involved and the equations applicable to the scenario. Some have provided calculations for acceleration and time, but discrepancies in results have led to further questioning and exploration of the problem.

Contextual Notes

There is a noted confusion regarding the large values involved in the calculations, and participants are grappling with the implications of their results. The discussion reflects a mix of attempts to apply formulas and check assumptions about the physics involved.

balllla
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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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