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Speed of bullet

  1. Sep 27, 2004 #1
    The speed of a bullet as it travles down the barrel of a rifle towards the opening is given be the expression v=(-5.0*10^7)t^2 + (3.0*10^5)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.

    (a) determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel.

    I multiplied everything by t to get the position equation: x=(-5.0*10^7)t^3 + (3.0*10^5)t^2

    I divided everything by t to get the acceleration equation: a=(-5.0*10^7)t + (3.0*10^5)

    is this thougth process correct?

    (b) determine the length of time the bullet is accelerated.
    you dont know the length of the barrel so is this possible?

    (c) Find the speed at which the bullet leaves the barrel
    based on question b

    (d) what is the length of the barrel
    based on question b as well

    in need of need hits
  2. jcsd
  3. Sep 27, 2004 #2


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    Differentiate to find the acceleration and integrate to find the position.
  4. Sep 27, 2004 #3
    what does differentiate mean?
  5. Sep 27, 2004 #4


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    Differentiate means finding the derivative. I assumed from the stated problem that you probably have some calculus experience. If not then you may have to resort to graphing and finding the slope of the curve at several points to make a graph of acceleration.
  6. Sep 27, 2004 #5
    for question B, I need to find the time, how would I do that?
  7. Sep 28, 2004 #6
    [tex]a= 300000-100000000t[/tex]

    [tex]x=.9 meters [/tex]
  8. Sep 28, 2004 #7


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    For that one you have to integrate!
  9. Sep 28, 2004 #8
    for the velocity question just plug in when you solved for time
  10. Sep 28, 2004 #9
    I just found the derivative of the V for a and integrated V for x
  11. Sep 28, 2004 #10
    how do you know acceleration is zero?
  12. Sep 28, 2004 #11
    Easy way to find derivitave
    take each chuck and do dervitiave of [tex]cx^n = ncx^{n-1}[/tex]
    to integrate
    [tex] bx^n = (n+1)x= c/(n+1)x^{n+1} [/tex]

    where n is power
    c is orignial coeffiecnt
    x is variable
  13. Sep 28, 2004 #12
    You gave that to me in the problem
  14. Sep 28, 2004 #13
    oh, that's right!! So the speed of the bullet would just be m/s.....9m/.003s?
  15. Sep 28, 2004 #14
    I just found velocity for it usuing the original equation
    you gave me
    which gave a velocity of 450 m/s
  16. Sep 28, 2004 #15
    wait...all I have to do is sub .003 into the original velocity equation to get 450m.s right?
  17. Sep 28, 2004 #16
  18. Sep 28, 2004 #17
    btw the x distance i got was 0.9 meters not 9 meters
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