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Homework Help: Still Friction

  1. Mar 21, 2007 #1
    So I posted this earlier and it got to a point where differentiation is neccessary. I am a little familiar with differentiation, but not to the point where I know how to apply the concepts I have recently learned in Calculus. i would greatly appreciate somebody walking me through the process...it should be fairly simple in this particular problem. Again, this problem is for my own personal practice, not for a class, and I am very interested in finding out how to apply these concepts.
    Thanks~Casey

    Original Post:
    1. The problem statement, all variables and given/known data
    A 1000kg Boat is traveling 90km/h when its engine is cut. The magnitude of the frictional force fk is proportional to the boat's speed v: fk=70v, where v is in m/s and fk is in newtons. Find the time required for the boat to slow to 45 km/h.



    2. Relevant equations
    Newton's Second
    V^2=Vo^2+2a(X-Xo)
    X-Xo=VoT+1/2at^2
    V=Vo+aT

    3. The attempt at a solution
    Vo=25m/s
    V=12.5m/s
    fk=70v=1750N

    I drew a FBD and it seems that since the engine was cut, there is only fk in the x direction. Thus, fk=ma--->1750=-1000a-->a=-1.75
    Then I used V=Vo+at---> t=(V-Vo)/a
    -->t=(12.5-25)/-1.75=7.1
    But this is not correct....9.9seconds is the correct solution.
    Any advice is appreciated.
    ~Casey

    ...It was pointed out that "a" is not constant. But I am not sure where to go from here as I have only dealt with problems dealing with constant acceleration.....what am I differentiating? I am not sure of the equation...or how to derive one. hollah.
     
  2. jcsd
  3. Mar 21, 2007 #2
    Yeah. The retardation is not constant since the magnitude of the frictional force depends on the speed of the boat.

    So, at a particular instant of time, assume the speed to be v. At that moment, the force acting on it is kv in the -x direction (k=70). Which means that the instantaneous acceleration is v(k/m).

    This means, [tex]\frac{dv}{dt}=\frac{kv}{m}[/tex]
    Integrating this equation, you get the expression, [tex]ln(v)=\frac{kt}{m}+c[/tex] where c is the constant of integration.

    Now, at t=0, the engine was cut and the boat had a speed of 90km/h.
    Putting that into the equation, you get c=ln(90).
    So, your final expression is:
    [tex]ln(v)=\frac{kt}{m}+ln(90)[/tex], where ln is natural logarithm.

    Using this, put v=45 and solve for t to get the time required.
     
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