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Physics kinematics calculus question

  1. Jun 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Initial acceleration of a particle moving in a straight line is a° and initial velocity is zero.
    The acceleration reduces continuously to half in every t° seconds. What will be the terminal velocity(v)?


    2. Relevant equations
    [itex]\int[/itex]adt=[itex]\int[/itex]dv


    3. The attempt at a solution
    [itex]^{∞}_{0}[/itex][itex]\int[/itex]adt=[itex]^{v}_{0}[/itex][itex]\int[/itex]dv
    (velocity will become terminal only when accelaration becomes 0 and since acceleration is becoming half after every time interval, it will become zero at infinity)
    But to use this equation I need to find an expression for acceleration in terms of time. How should i do it? Should i use geometric progression?
     
    Last edited: Jun 28, 2012
  2. jcsd
  3. Jun 28, 2012 #2

    Simon Bridge

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    a(t) is a decaying exponential - the time to reduce by half is called the "half life".
    You can derive the formula if you like ... start with da/dt=λa knowing that a(t°)=a°/2

    You should find that the velocity-time graph is asymptotic to some vmax.
     
    Last edited: Jun 28, 2012
  4. Jun 28, 2012 #3
    Hi shiv99, welcome to PF :smile:

    The final velocity of the particle for the first t seconds is given as [itex]v_1=at[/itex], for the next t seconds is, [itex]v_2= at+at/2[/itex] and so on.

    So yes, you can apply the infinite geometric sum for these.
     
  5. Jun 28, 2012 #4

    Curious3141

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    This would only be correct if the particle's acceleration was halving *discontinuously* every t0 seconds.

    The correct approach in this case is to set up a differential equation for a, i.e. [itex]\frac{da}{dt} = -{\lambda}a[/itex] and solve it. Basically, this is the same equation as in exponential radioactive decay.

    After integrating (use the conditions given to determine the appropriate bounds), you'll get a closed form expression for a(t).

    Now [itex]v(t) = \int_0^t a(t)dt[/itex].

    Solve for v(t) and find [itex]\lim_{t \rightarrow \infty} v(t)[/itex] to determine the terminal velocity. The answer should have [itex]\ln 2[/itex] in it.
     
  6. Jun 28, 2012 #5
    Ahh, my bad. :blushing:
     
  7. Jun 28, 2012 #6

    Simon Bridge

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    @Infinitum: you had me panicking for a bit there :)
     
  8. Jun 28, 2012 #7
    Here what will be the value of lambda for this ques. Plz help because this(exponential radioactive decay)is a new concept for me
     
  9. Jun 28, 2012 #8

    Simon Bridge

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    Lambda is a dummy constant - you find out what it is after you solve the differential equation by using the fact that you know one of the solutions, vis: a(t°)=a°/2 ... you'll also get a constant of integration which you find from a(0).
     
  10. Jun 28, 2012 #9
    But how do you use it find a(t)
     
  11. Jun 28, 2012 #10

    Simon Bridge

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    You find a(t) by solving the differential equation provided.

    One last time: a(t) is the solution to: [itex]\frac{da}{dt} = \lambda a[/itex] knowing that [itex]a(t=0)=a_0[/itex] and [itex]a(t=T_{1/2})=a_0/2[/itex]

    Or you can google "exponential decay" or "radioactive decay" and use the solution provided.
     
  12. Jun 28, 2012 #11

    Curious3141

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    Simon has already covered the essentials. You need to express [itex]\lambda[/itex] in terms of [itex]t_0[/itex] by solving the differential equation, then using the "half-life" condition stated.
     
  13. Jun 28, 2012 #12

    Simon Bridge

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    I'm allowing that Shiv may not know how to solve differential equations ... though just looking up the solution and adapting it to this specific case is also valid. It wouldn't be the first time someone was doing a problem without the broader knowledgebase the rest of us take for granted. Nothing wrong with that as such - it shows the ambition and courage of the self taught.
     
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