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Acceleration Due to Gravity

  1. Feb 5, 2016 #1
    1. The problem statement, all variables and given/known data
    An elevator moving downward at 4.00m/s experiences an upward acceleration of 2.00m/s^2 for 1.80seconds. What is its velocity at the end of the acceleration and how far has it travelled?

    downward acceleration=-9.8m/s^2, upward acceleration=2.00m/s^2. time=1.80seconds. Initial velocity=4.00m/s?
    final velocity=? displacement=?
    2. Relevant equations
    One of the big 5 kinematics equations, I'm not sure which one.

    3. The attempt at a solution
    I tried to solve this using vf=vi+a(t), but it didn't work. The correct answer for final velocity is 0.400m/s[down], and for distance travelled is 3.96m. But, I don't know how to get to this answer. I think what I plugged in for the initial velocity(4.00m/s) might be wrong
     
  2. jcsd
  3. Feb 5, 2016 #2
    Since the lift is also under action of gravity, you need to take relative acceleration.
     
  4. Feb 5, 2016 #3
    What does that mean? It's under the force of gravity when it's accelerating up.
     
  5. Feb 5, 2016 #4
    Okay, so in your equation vf=vi+a*t, take care of the signs of velocity and acceleration. If you take downward direction to be positive, a will be negative, vi will be positive. Now calculate vf.
    My bad, relative not required.
     
  6. Feb 5, 2016 #5

    SteamKing

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    We don't know what happened in your calculations because you didn't show them.

    Remember, the elevator is moving down initially at 4.00 m/s, while the acceleration is acting upward at 2.00 m/s2. What does this suggests happens to the final velocity of the elevator? How would you indicate this mathematically?

    It takes a different SUVAT equation to find the distance the elevator travels while the velocity is changing. Which equation contains distance, velocity, acceleration, and time?
     
  7. Feb 5, 2016 #6
    How is it possible that the elevator is moving down while the acceleration is in the upwards direction. I thought that the question described 2 different periods of motion-one where the elevator is accelerating downwards and another one when the elevator momentarily accelerated upwards, and it asks for the final velocity at the end of the second period of motion(when the elevator is accelerating upwards). That's what I think.
     
  8. Feb 5, 2016 #7
    What is the vi value? Is the one I wrote correct?
     
  9. Feb 5, 2016 #8

    SteamKing

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    Who knows? Perhaps the car is snagged on something which is acting like a brake to slow it down. That's not important to finding out what happens to the velocity of the elevator.
    It doesn't appear that's what the answers describe.
    vi is 4.00 m/s, down. Direction is important here.
     
  10. Feb 5, 2016 #9
    I tried it again- I think i forgot to plug in the vectors with the proper signs(negative for down, positive for up), because I got the right answer for final velocity. I then used vf^2=vi^2+2a(displacement)--- (-0.4)^2=4.0^+2(2)(displacement)-- if you solve that, you get displacement=-3.96m. Did I do something wrong becuse the distance travelled is +3.96m as indicated by the answer to the question.
     
  11. Feb 5, 2016 #10

    SteamKing

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    You're using the right numbers, but your arithmetic seems a bit dodgy.

    For the velocity, we're given that the elevator is traveling at 4.00 m/s, down. Let's say down in negative, so vi = -4.00 m/s. Whilst the elevator is moving down at this velocity, a temporary acceleration of 2.00 m/s2 upward occurs, so a = +2.00 m/s2, which lasts for t = 1.80 sec.

    What vf do you calculate if you plug these numbers into vf = vi + a ⋅ t ?

    What about when you plug them into vf2 = vi2 + 2 ⋅ a ⋅ s ?
     
  12. Feb 5, 2016 #11
    For the first formula, I calculated -0.4 for vf which is 0.4m/s[down], and for the second formula you gave, I'm not sure what the variable 's' represents-- is it displacement because that is the variable that should go after 'a' in that formula, I believe.
     
  13. Feb 5, 2016 #12

    SteamKing

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    s is often used for displacement.

    vf is correct, BTW.
     
  14. Feb 5, 2016 #13
    I got -3.96 for displacement using the second formula. I know that displacement=df-di. but I think di=0, so that would make df=-3.96, which is equal to the distance travelled- I still don't understand why my answer is negative.
     
  15. Feb 5, 2016 #14

    SteamKing

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    Well, the initial velocity of the elevator is 4.00 m/s going down, and the final velocity is 0.40 m/s going down, so the elevator has been going down during the entire interval of 1.80 sec.
     
  16. Feb 5, 2016 #15
    Oh, so it just indicates that the distance is travelled downwards.
     
  17. Feb 5, 2016 #16

    SteamKing

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    Yes.
     
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