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Physics mechanics kinematics question?

  1. Sep 22, 2008 #1
    The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v=(-5.00x10^7)(t^squared)+(3.00x10^3)(t)... v=m/s, and t=s. Acceleration of the bullet just as it leaves the barrel is 0m/s^squared.

    Questions is: How do i find the acceleration and position of the bullet as a function of time when the bullet is in the barrel?
    length of time the bullet is accelerated?
    the speed at which the bullet leaves the barrel?
    what is the length of the barrel?

    that is the complete question: however what i really want to know is what are the actual givens so i can carry out the kinematics? I can't figure out what to take out from what is given to me.
  2. jcsd
  3. Sep 22, 2008 #2


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    Acceleration is the derivative of the velocity function, it is just the rate of change of velocity. See if you can work from there.
  4. Sep 22, 2008 #3
    k i got
    a=-(10.00x10^7m/s^3)(t)+(3.00x10^5m/s^2) as the derivative
  5. Sep 22, 2008 #4
    The answers, respectively, are:
    a: x=-(1.67x10^7m/s^3)t^3 + (1.50x10^5m/s^2)t^2
    b: 3.00x10^-3s

    I need help finding out how to solve this problem, so far I've got a=-(10.00x10^7)(t)+(3.00x10^5) as the derivative (don't understand how it's 1.67...)
    and that the acceleration of the bullet just as it leaves the barrel is 0.

    ok for b i figured out that when the bullet leaves the barrel the acceleration is 0, so if i set the derivative of the function given to equal to 0 i will have the time, i did that and got .003s which is good.
    Last edited: Sep 22, 2008
  6. Sep 22, 2008 #5


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    You've been given the Velocity function as a function of time.

    As already noted the derivative of the velocity function is the acceleration function, which apparently is still time variant.

    If velocity is the derivative of position, then ... you simply integrate the Velocity function to derive the position function with respect to time.

    Now the length of the barrel requires thinking about the problem a bit. For instance when will acceleration be 0 from the propelling gas pressure in the barrel? (Hint: Maybe when there is no more barrel?)

    And at what time will that be? Won't that time plugged in the velocity equation then describe its velocity of exit and likewise plugging into the position equation yield the barrel length?
  7. Sep 22, 2008 #6
    I manage to understand everything except on how to get the position function with respect to time.
  8. Sep 22, 2008 #7


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    You know that velocity is dx/dt.

    What then is the position? Isn't it the product of velocity and time? And if you add up all the instantaneous time segments and multiply each segment by the velocity at that instant, you are left with little areas that represent this velocity*time product. The sum of those areas then from any given time to any other given time is the summation of all the infinitesimal area elements and this sum is the area under the v-curve and hence your distance.

    So if you take the integral of V, from T=0 to T=t, then you should get as a result X (the area under the V curve) as a function of "t".

    Is taking the Integral of V your problem?
    Last edited: Sep 22, 2008
  9. Sep 22, 2008 #8
    Yes it's taking the integral my problem...:shy:
  10. Sep 22, 2008 #9


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    Easily solved then.

    Your equation: v=(-5.00x10^7)(t^squared)+(3.00x10^3)(t)

    Let's rewrite that as:

    [tex] v_{(t)} = \frac{dv}{dt} = A*t^2 + B*t [/tex]

    Where A and B are constants. Integrating that then

    [tex] x_{(t)} = \int dv = \int (A*t^2 + B*t) dt = \frac{A*t^3}{3} + \frac{B*t^2}{2} + C [/tex]

    C is a constant of integration that we know is 0, because at t=0, x=0

    So ...:

    [tex] x_{(t)} = \frac{A*t^3}{3} + \frac{B*t^2}{2} [/tex]
  11. Sep 22, 2008 #10
    Woah thanks a lot for your help.
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