Finding time it takes for electron to travel a certain distance

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SUMMARY

The discussion focuses on calculating the time it takes for an electron to travel a distance of 3 cm while accelerating uniformly from an initial speed of 9 x 103 m/s to a final speed of 6 x 106 m/s. The relevant equation used is d = 1/2(v0 + v)t, where d is the distance, v0 is the initial velocity, and v is the final velocity. The solution involves converting the distance from centimeters to meters and simplifying the equation to find time, ultimately leading to the calculation of average speed and time taken.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = 1/2(v0 + v)t
  • Knowledge of unit conversion between centimeters and meters
  • Basic algebra skills for manipulating equations
  • Familiarity with average speed calculations
NEXT STEPS
  • Study the derivation and applications of kinematic equations in physics
  • Learn about uniform acceleration and its implications in motion problems
  • Practice unit conversion techniques for various physical quantities
  • Explore average speed and velocity concepts in different contexts
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Students studying physics, particularly those focusing on kinematics and motion, as well as educators looking for examples of uniform acceleration problems.

GoGoGadget
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Homework Statement


Over a distance of 3 cm an electron accelerates uniformly from a speed of 9 x 103 m/s to a speed of 6 x 106 m/s. How long does it take the electron to cross this distance?


Homework Equations



d = 1/2(v0 + v)t


The Attempt at a Solution



Find t:

d = 3 cm /10 cm = 0.3 m
2 d / v0 + v = t

2 (0.3 m) / (9 x 103 m/s + 6 x 106 m/s) = t

This is where I got a little lost, especially with trying to do it on paper as I'm not able to use a calculator for the test.
 
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GoGoGadget said:
d = 3 cm /10 cm = 0.3 m
Umm.. how many cm to the m?
2 (0.3 m) / (9 x 103 m/s + 6 x 106 m/s) = t

This is where I got a little lost, especially with trying to do it on paper as I'm not able to use a calculator for the test.
First, there's a simple cancellation out of a factor. If you then write out the denominator in fixed point, what do you notice?
 
Average speed is V = 600900000/2 =300450000 cm
time is 3 /V = 3/30045 (*10000) can you do 3.00000/30045 on paper? and then 0.00000000...
 
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