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Motion in One Dimension (Using Calculus)

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v = (-5.95 multiplied by 107) t 2 + (2.45 multiplied by 105) t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero.

    (b) Determine the length of time the bullet is accelerated.
    (c) Find the speed at which the bullet leaves the barrel.
    (d) What is the length of the barrel?


    2. Relevant equations
    Taking the derivative of the given equation, I got the function for acceleration, and integrating the given function, I got the position function.
    a(t)=-1.19*10^8t+2.45*10^5
    x(t)=-1.98*10^7t^3+1.23*10^5t^2

    3. The attempt at a solution
    (b)Using the acceleration formula I derived, I plugged in 0 because the given info says that the acceleration of the bullet just as it leaves the barrel is 0, and got an answer of .002m/s^2. The Webassign is telling me that the answer is within 10% of the correct answer, but I'm stuck.
    (c&d) I don't even think there is enough information to solve for either of these.
     
  2. jcsd
  3. Sep 6, 2008 #2

    LowlyPion

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    Welcome to PF.

    Just to be sure I have it right:

    [tex]v = 2.45 * 10^5 * t - 5.95 * 10^7 * t^2[/tex]
     
  4. Sep 6, 2008 #3
    Oh, sorry, I forgot to adjust it in the given information.

    v(t) = -5.95*10^7*t^2 + 2.45*10^5*t
     
  5. Sep 6, 2008 #4

    LowlyPion

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    a(t)=-1.19*10^8t+2.45*10^5

    When you set a(t) = 0, you get the time that a is 0.

    Your answer is .002 m/s^2 ?
     
  6. Sep 6, 2008 #5

    LowlyPion

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    Using significant digits digits more formally, rather than the .002 you submitted, I would suggest a more precise 2.06 x 10^-3 to be consistent with the form of your givens.
     
  7. Sep 6, 2008 #6
    I didn't put 0 in as the value for t, I made the function equal to 0, because I believe that the problem stated that the acceleration of the bullet was 0, but I'm not exactly sure. Actually, reading over the problem again, it's asking for the amount of time that the bullet was accelerated, so I guess it had to be at one point, obviously.

    I just don't see how there is enough information to solve for the answers it wants.

    I also tried putting the value as 2.06 x 10^-3 and was also wrong.
     
  8. Sep 6, 2008 #7

    LowlyPion

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    I understand. But when you make the acceleration function a(t) = 0, that yields a value of time. It's units are in seconds are they not? Not units of acceleration - m/s^2?

    Isn't that then the time you are looking for with which to solve for the other values?
     
  9. Sep 7, 2008 #8
    Oh, okay. Yes, I see what you're saying. The number is right, but yes, the units should be in seconds. I talked it over with a friend of mine who took AP Physics last year, and he agreed with me that the time that was yielded from setting a(t)=0 is the amount of time the bullet was in the barrel, which is also the amount of time it was accelerated, but for some reason, the answer isn't being accepted. Any other ideas? =/
     
  10. Sep 7, 2008 #9

    LowlyPion

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    From the statement that I have of the problem and reviewing your integral and derivative that's the time I would use to solve for V and the x from 0, i.e. the barrel length.

    Is there a part (a) to the question?
     
  11. Sep 7, 2008 #10
    Alright, I retried doing the problem. For part (b), instead of putting .0021, I put .00206, and it gave me credit for the problem. Using that time value instead of the first one, all the rest of the answers were correct. Thanks for you help though, I hope you didn't get as frustrated as I did.
     
  12. Sep 7, 2008 #11

    LowlyPion

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    No frustration at all for me. Glad to see it resolved. Congrats to you for staying after it.
     
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