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Initial Acceleration

  1. Sep 9, 2007 #1
    1. The problem statement, all variables and given/known data

    A truck drives along a straight line, which we will call the x-axis with the positive direction to the right. The equation for the turtle's position as a function of time is:

    x(t) = 50.0cm + (2.00cm/s)t - (.0625cm/s^2)t^2

    Find initial velocity:
    Find initial acceleration:
    Find initial position:


    2. Relevant equations
    [​IMG]


    3. The attempt at a solution

    I found that initial velocity is 2.00cm/s, as the equation states that V0 is initial velocity.

    I found initial position, by knowing that if time was zero, then the obvious would be to have it at 50cm.:bugeye:

    The acceleration is giving me the problem, I would think that it was .0625cm/s^2, but it is wrong. Then I would guess that since the initial velocity is 2.00cm/s then, acceleration at that point in time might be the same...is this right?

    I found this equation for instantaneous acceleration in which a= delta vx / delta t.

    So this would be

    2.00cm/s / 0 = 0 acceleration? :confused:
     
  2. jcsd
  3. Sep 9, 2007 #2

    G01

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    What about the 1/2 in this equation?:

    [tex]x(t)=x_0+v_0t+1/2at^2[/tex]


    Also, remember that you have -.0625, not +.0625

    EDIT: Perfect Timing Doc Al! :smile:
     
    Last edited: Sep 9, 2007
  4. Sep 9, 2007 #3

    Doc Al

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    Why guess?

    Compare this specific equation:
    With the general equation for x(t) (the second equation):

    [​IMG]

    What must "a" equal?
     
  5. Sep 9, 2007 #4
    so initial acceleration must equal -.125 cm/s^2
     
    Last edited: Sep 9, 2007
  6. Sep 9, 2007 #5

    G01

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    Looks good!
     
  7. Sep 9, 2007 #6
    Ok, I understood that part, this is another part of the question:

    How long after starting does it take the truck to return to its starting point?

    I managed to find at what time the velocity of the truck would be zero, and it would be at 16s.

    Now I would setup it up like this,

    0 = 50.0cm + (2.00cm/s)t - (.0625 cm/s^2)t2

    -.0625t^2 + 2t + 50 = 0 <----I would solve for t using quadratic equation, although , I don't believe this is possible since they are cm/s and cm.s^2 etc.


    :confused:
     
  8. Sep 9, 2007 #7

    Doc Al

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    Looks good to me--solve it! The units for time will be seconds--no problem. (Note that each term in that equation for position has units of length--so everything is perfectly consistent.)
     
  9. Sep 9, 2007 #8
    Ok, this is what I got

    -2 +- sqrt (4-4(-.0625)(50)) / 2(-.0625)


    = -2+- 4.06 / -0.125


    = -16.48 and 48.48 seconds.

    since time cannot be negative, I answered 48.48 seconds, but I got it wrong.

    Can you help me find my mistake?

    Unless I add 16s (from when velocity was 0 at time 16s) with 48.48seconds, when the truck reaches same point ?

    I rechecked my calculation , updated it as above, and still got it wrong.
     
    Last edited: Sep 9, 2007
  10. Sep 9, 2007 #9

    learningphysics

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    Don't round that discriminant to 4.06. Carry everything as much as you can.
     
  11. Sep 9, 2007 #10
    -2 + - 4.062019202 / -.125 = -16.496 and 48.496

    does this seem right?

    sould I add the 16s when velocity was 0?

    would the answer the 64.496? or 48.496?

    A reminder of the question : "How long after starting does it take the truck to return to its starting point?"
     
  12. Sep 9, 2007 #11

    Doc Al

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    Oops... looks like they tricked us. :redface: It says return to the starting point. Assuming it starts at t = 0, the starting position is not x = 0. (You won't even need a quadratic to solve it now.)
     
  13. Sep 9, 2007 #12
    but if t = 0, then how could be solve for t? I am confused on what to use now :(
     
  14. Sep 9, 2007 #13

    Doc Al

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    t = 0 will give you the starting position; plug that position into your equation for x(t) and solve for t. (You know that one answer is t = 0; what's the other?)
     
  15. Sep 9, 2007 #14

    learningphysics

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    Those jerks!!!
     
  16. Sep 9, 2007 #15
    x(0) = 50.0cm + 2.00 cm/s (t) - 0.0625cm/s^2 (t)^2

    x(0) = 50.0cm + 2.00 cm/s (0) - 0.0625cm/s^2 (0)^2

    = 50.0cm

    50.0cm = 50.0cm + 2.00 cm/s (t) - 0.0625cm/s^2 (t)^2

    -.0625t^2 + 2t = 0

    t(-.0625t + 2) = 0
    t= 0 <---which we know
    and

    -.0625t +2 = 0

    t= 32seconds.

    so the time it will take to return to the starting point would be 32seconds. Am i right? :zzz:
     
  17. Sep 9, 2007 #16

    learningphysics

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    Yup. That's right.
     
  18. Sep 9, 2007 #17
    Sorry if I am bombarding you guys with questions.

    The next one got me too:

    At what time is the truck first time a distance of 10.0 cm from its starting point?

    This is what I did, assuming that it follows this format:

    10 = 50.0 2t - .0625t^2.

    t= -40s and 672 seconds.

    both are incorrect.
     
  19. Sep 9, 2007 #18

    Doc Al

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    Lets not get fooled again! The starting point is not at x=0! So what position are we talking about?
     
  20. Sep 9, 2007 #19
    well if the initial starting point is 50cm, then if I add 10cm more it would be ay 60cm, which I would have to solve for time.

    Do I solve this again for 60cm?
     
  21. Sep 9, 2007 #20

    Doc Al

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    Do it! ...
     
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