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anonymous

  • 4 years ago

An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 ms. What is the magnitude of this acceleration?

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  1. JamesJ
    • 4 years ago
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    What's the equation of motion for a body under constant acceleration?

  2. JamesJ
    • 4 years ago
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    Suppose v = initial velocity t = time a = constant acceleration Then what is the formula for the distance, d, the object moves in time t? d = ... what?

  3. Kainui
    • 4 years ago
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    What is the definition of acceleration? It's the change of velocity with respect to time. How would you put this in the form of an equation?

  4. JamesJ
    • 4 years ago
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    I'm going to change notation slightly, writing now u for initial velocity. Then we have: u = initial velocity t = time a = constant acceleration Write also v(t) = velocity of object at time t d(t) = position of object at time t Now, by definition, acceleration = (change in velocity)/(Time) hence \[ a = \frac{v(t) - u}{t} \] thus \[ v(t) - u = at \] and \[ v(t) = u + at \] Making sense so far?

  5. anonymous
    • 4 years ago
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    Yeah, I'm following

  6. JamesJ
    • 4 years ago
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    Now, what's d(t) ?

  7. anonymous
    • 4 years ago
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    The position?

  8. JamesJ
    • 4 years ago
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    yes, what's the formula for it?

  9. anonymous
    • 4 years ago
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    v1t + 1/2 a t^2?

  10. JamesJ
    • 4 years ago
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    Yes \[ d(t) = ut + \frac{1}{2}at^2 \] For your problem, you're told that the initial velocity is zero, t = 0.5 ms and d(0.5 ms) = 0.92 m Now calculate a.

  11. JamesJ
    • 4 years ago
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    *correction, d(0.5 ms) = 0.02 m, not 0.92.

  12. anonymous
    • 4 years ago
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    would I try and get a by its self?

  13. JamesJ
    • 4 years ago
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    Thus u = 0 m/s t = 0.005 s d = 0.02 m Substitute that into the equation above and solve for a

  14. anonymous
    • 4 years ago
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    .02 = 0(.005) + .5(a)(.005)^2?

  15. JamesJ
    • 4 years ago
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    yes

  16. anonymous
    • 4 years ago
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    So if a=(v(t)−u)/t, that would equal v(t)/t, correct?

  17. JamesJ
    • 4 years ago
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    yes

  18. JamesJ
    • 4 years ago
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    but here velocity is not constant so you have to be very careful with that formula

  19. anonymous
    • 4 years ago
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    But the velocity equation is dependent on the acceleration equation: v(t)=u+at

  20. JamesJ
    • 4 years ago
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    Yes, so here because acceleration is constant, we're good. I.e., the average acceleration is equal to the constant acceleration.

  21. anonymous
    • 4 years ago
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    But the closest we got to solving for a is: .02 = 0(.005) + .5(a)(.005)^2 and v(t) needs a, v(t)=u+at I'm not getting this part, how do we solve for either one?

  22. anonymous
    • 4 years ago
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    ?

  23. JamesJ
    • 4 years ago
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    The question only asks you to find the acceleration. Hence you only need one equation and the one that is germane here--because it uses the information of the problem--is the the one into which you've already substituted.

  24. JamesJ
    • 4 years ago
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    Hence you have: .02 = 0(.005) + .5(a)(.005)^2 This already simplifies to \[ \frac{0.005^2}{2}a = 0.02 \] Now solve for \( a \).

  25. anonymous
    • 4 years ago
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    A = .02(2/.005^2) = .04/.005^2?

  26. JamesJ
    • 4 years ago
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    Evaluate it. You can't leave the answer in that form.

  27. anonymous
    • 4 years ago
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    1600?

  28. anonymous
    • 4 years ago
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    What about the units though, is that in meters?

  29. JamesJ
    • 4 years ago
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    We were (or at least I was) careful to convert all the units into SI units before we started calculating time = seconds distance = meters velocity = meters/sec = m/s acceleration = m/s^2

  30. anonymous
    • 4 years ago
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    Ah, so its 1.6 km/s^2?

  31. JamesJ
    • 4 years ago
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    yes

  32. anonymous
    • 4 years ago
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    Awesome! Thanks for all your help!

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