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What is happening in this step, kinematics problem

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data
    a particle travels along a straight line and the acceleration is given by
    a = 30 - 0.2v
    determine the time when the velocity of the particle is v = 30


    2. Relevant equations

    [tex]\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b) [/tex]


    3. The attempt at a solution
    This is the solution from the textbook. The problem is that my math class is behind schedule so I haven't learned integrals yet. The only knowledge I have on the matter is from what was briefly covered in my dynamics class. So I need an explanation of what is happening is happening in this solution.
    [tex] dt = \frac{dv}{a} [/tex]
    [tex]\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v} [/tex]
    [tex]t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0 [/tex]
    I understand everything upto this point but I have no idea what is happening in the next step.
    [tex] t = 5 \ln \frac {30}{30-0.2v} [/tex]
    [tex] t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s [/tex]
    several questions, first where did the 30 in the numerator come from? why did the minus sign before [tex] \frac{-1}{0.2} = -5 [/tex] go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.
     
  2. jcsd
  3. Jan 15, 2013 #2

    Mark44

    Staff: Mentor

    Typo above - it should be dx in the first integral.
    The right side above is -5[ln(30 - .2v) - ln(30)] = 5[ln(30) - ln(30 - .2v)]
    $$= 5 ln\frac{30}{30 - .2v}$$
    They're just using a property of logs; namely that ln(a/b) = ln(a) - ln(b).
     
  4. Jan 15, 2013 #3
    did your post cut off or something?
     
  5. Jan 15, 2013 #4

    Dick

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    No, I think it's all there. Look at the part before the last quote. That's answering the question of why 5 instead of -5 and the manipulation of the logs stuff.
     
  6. Jan 15, 2013 #5
    well i managed to figure it out when I found out that the log property was used. I don't understand the point of putting the variable in the denominator instead of simply writing
    [tex] \ln \frac{30-0.2v}{30} [/tex]
    why make things unnecessarily complicated just to get rid of a negative sign.
    Anyways thanks for the help.
     
  7. Jan 15, 2013 #6

    Dick

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    Yes, there was no NEED to get rid of the negative. They just chose to write it that way.
     
  8. Jan 16, 2013 #7

    Ray Vickson

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    There is some vital information missing from the problem statement: what is the initial velocity v0 at t = 0? For example, if v0 = 50 we cannot have v(t) = 30 for t > 0; the solution has t < 0; if v0 > 150 there is no time where v(t) = 30.
     
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