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Need to calculate acceleration from total distance travelled.

  1. Jan 22, 2012 #1
    Two objects move with initial velocity -7.9 m/s, final velocity 15.8 m/s, and constant accelerations.
    (a) The first object has displacement 17.5 m. Find its acceleration.
    Answer is 5.35 m/s^2 <-- I got this right.

    (b) The second object travels a total distance of 23.0 m. Find its acceleration.
    I am having trouble with this one.
    So far I tried 8.14m/s^2 and 4.07 m/s^2. Webassign says it is wrong.
    I don't know what I am doing wrong. :(
     
  2. jcsd
  3. Jan 22, 2012 #2

    tiny-tim

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    welcome to pf!

    hi monac! welcome to pf! :wink:
    show us how you got those, and then we can see where you've gone wrong :smile:
     
  4. Jan 22, 2012 #3
    a = change in v / change in t
    t = 23m/7.9 m/s = 2.91 s <-- doesn't make sense because there are two different velocities so this approach is wrong.
    a = 15.8 + 7.9 / 2.91 s
    a = 8.14 m/s^2


    d = v1+v2/2*t
    23 = (15.8+-7.9)/2
    solved for t
    t = 5.82 s
    plug this into vf = vi + at
    15.8 + 7.9/ 5.82s
    a = 4.07 m/s^2
     
  5. Jan 22, 2012 #4
    Thank you! I never used Physics Forums before, it is my first time.
     
  6. Jan 22, 2012 #5

    tiny-tim

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    hi monac! :smile:
    correct! :smile: (ie, it is wrong! :redface:)

    you need a constant acceleration equation with u v a and s (not u v a and t) …

    if you don't know one, look up constant acceleration in the pf library or wikipedia :wink:
     
  7. Jan 22, 2012 #6
    When I plug into v^2 = u^2 + 2as
    I am getting 4.07 m/s^2 again which is the wrong answer.
    And that's the only equation without a t.
     
  8. Jan 22, 2012 #7

    tiny-tim

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    hmm … yes, that's exactly the correct equation :smile:

    it's also exactly the answer i get :confused:

    (and just to check, it's 17.5/23.0 times the answer to the first part)

    the book must be wrong :redface:

    oh wait a mo …

    does the question say "total distance" instead of "displacement", or is that just you? …

    the total distance would be adding the negative to the positive, instead of subtracting :wink:

    goodnight! :zzz:​
     
  9. Jan 22, 2012 #8
    It's the total distance not the displacement
    That's why I am having trouble with this problem
    What do you mean by

    "total distance would be adding the negative to the positive, instead of subtracting"?
     
  10. Jan 22, 2012 #9
    try splitting part b into two parts, with zero velocity as the dividing point
     
  11. Jan 22, 2012 #10
    so you are saying v0 = 0 m/s
    vf = 15.8 m/s ?
    Then what would be delta x?
    15.8^2 - 0^2 / 2*23?
    Do you think that's right?
     
  12. Jan 22, 2012 #11
    hmm no wait, basically the 23 meters is divided into a distance traveled backwards and a distance traveled forwards. 23 meters is the total distance traveled, but what wikipedia wants is the displacement of the object.

    when wikipedia uses s, it means x - x0, which is the displacement, that's not the value we're given, we're given the total distance traveled.

    On wikipedia it also shows us that s = [itex]\frac{(u + v)t}{2}[/itex]

    and this is the displacement for our situation, except we don't know the displacement, we only know the total distance traveled. So to get the total distance traveled, we pretend that our initial velocity is positive instead of negative.
     
  13. Jan 22, 2012 #12
    So I understand that our initial velocity should be positive, then ... what equation do I plug it into? all equations have a t in them ... which is unknown.
     
  14. Jan 22, 2012 #13
    but when s is total distance, it's not unknown, right?

    total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex] basically
     
  15. Jan 22, 2012 #14
    so
    2s/u+v t
    2*23/7.9+15.8 = t
    1.94 s = t

    vf - vi / t = a
    15.8 - 7.9 / 1.94 = a
    a = 4.07 m/s^2 <-- That's the wrong answer I got in the first place :(
    This is disappointing.
     
  16. Jan 22, 2012 #15
    well when you are no longer worried about the distinction between total distance traveled and displacement, then you no longer have the absolute value symbols there.
     
  17. Jan 22, 2012 #16
    You no longer have the absolute value symbols where? in the second part of my solution?
     
  18. Jan 22, 2012 #17
    if s = [itex]\frac{(u + v)t}{2}[/itex], then total distance traveled = [itex]\frac{(|u| + |v|)t}{2}[/itex]

    but v = u + at is still valid, it's not |v| = |u| + at, since we aren't worried about making a distinction between total distance traveled
     
  19. Jan 23, 2012 #18

    tiny-tim

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    (just got up :zzz:)

    As SHISHKABOB :smile: says …
    … so you need to use three values of the speed instead of the usual two (in v2 = u2 + 2as) …

    u = vi, v = 0, then u = 0 and v = vf,

    and find separate values of |s| for the vi to 0 part, and for the 0 to vf part :wink:
     
  20. Jan 23, 2012 #19
    your process is still giving me 4.07 m/s^2 which is the wrong answer :(
     
  21. Jan 23, 2012 #20
    how do I get the s when I don't know the t?
     
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