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Physics with Calculus problem

  1. Jan 14, 2010 #1
    The position of the front bumper of a test car under microprocessor control is given by x(t)=2.17m + (4.80m/s^2)t^2-(0.100m/s^6)t^6
    Find its position at the first instant when the car has zero velocity.
    Find its acceleration at the first instant when the car has zero velocity
    Find its position at the second instant when the car has zero velocity
    Find its acceleration at the second instant when the car has zero velocity
    Please anyone, I need help with this, I'm not getting the correct answer!! Thanks
     
  2. jcsd
  3. Jan 14, 2010 #2

    rock.freak667

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    velocity=x'(t)
    acceleration=v'(t)=x''(t)

    you will need to use these identities.
     
  4. Jan 14, 2010 #3

    HallsofIvy

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    Differentiate to find the velocity function. Set that equal to 0 and solve for t. There apparently will be more than one solution. Put the smallest positive solution into the x(t) function and evaluate.

    Differentiate again to find the acceleration function. Put the t you found above into that acceleration function and evaluate.

    This is why I said there "apparently will be more than one solution". Set t equal to the next larger solution and evaluate x(t) for that t.

    Put that new value of t into the acceleration function and evaluate. If you have done these things and are not getting the correct answer, show what you have done.
     
  5. Jan 17, 2010 #4
    x(t) = 0 + 2(4.80m/s^2)t - 6(0.100m/s^6)t^5
    dx/dt = 2(4.80)t - 6(0.100)t^5=0
    I keep getting t=16, but thats not right, please help me solve!!!!! please!!!!!!
     
  6. Jan 17, 2010 #5

    tiny-tim

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    Welcome to PF!

    Hi dovec! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    no you don't! :wink: put that equal to 0, and divide by t, and you get … ? :smile:
     
  7. Jan 18, 2010 #6

    HallsofIvy

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    This is not dx/dt, it's just x(t) again, with the initial "0" dropped!

    If [itex]x(t)= 9.6 t- .6 t^5[/itex], then [itex]dx/dt= 9.6- 3.0t^4[/itex]
    Set that equal to 0 and you get [itex]t^4= 9.6/3.0= 3.2[/itex]. Take the fourth root of 3.2. It is definitely NOT 16!
     
  8. Jan 20, 2010 #7
    ok thanks alot guys, I think I finally figured it out.
     
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