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Boat Friction Physics Problem

  1. Mar 20, 2007 #1
    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
     
    Last edited: Mar 20, 2007
  2. jcsd
  3. Mar 20, 2007 #2
    You have assumed that the frictional force is constant when it is not.
     
  4. Mar 20, 2007 #3
    The frictional force changes as the velocity of the boat slows down. Your equations seem to be ignoring this fact.
     
  5. Mar 20, 2007 #4
    I had contemplated that, but was not sure how to deal with it...so I guess I just ignored it! I have not dealt with a problem like this before. I just picked it randomly out of the text for practice. I could use a hint on this one. It's kind of making we crazy. If fk is constantly shrinking, assuming that deceleration is constant, what can I do to alter my equations?
     
  6. Mar 20, 2007 #5
    First, find the acceleration a of the boat in terms of v. Then, use one of the equations relating a and v and solve for v. You will now have an equation for v in terms of t which you can then use to solve the problem.
     
  7. Mar 20, 2007 #6
    I am sorry, I don't follow. a=(V-Vo)/t and then....? I always have three unknowns. a,t, and displacement of x. I am given Vo and V.....
     
  8. Mar 20, 2007 #7
    You are given the velocity when the engines are cut, which you can consider the initial velocity, the final velocity of the boat and the friction on the boat. The unknown is the time when the boat reaches the final velocity.

    Do you care about the displacement of the boat? I think not. Do you care about the boats acceleration? Yes. How would you go about finding it?
     
  9. Mar 20, 2007 #8
    I am given that the initial V=25m/s and the final is 12.5m/s. there is only one equation that does not involve x. V=Vo+at
    Also fk=ma.
    a=(V-Vo)/t and a=fk/m. If the V is changing ....I am lost.
     
  10. Mar 20, 2007 #9
    Good. You know what fk is in terms of v. Plug that back into the second equation for a. Now you have two equations for a. What should you do with them?
     
  11. Mar 20, 2007 #10
    This isn't working out. fk=70v but v is changing... a=fk/m
    and a=(V-Vo)/t...I can set (V-Vo)/t=fk/m but I still do not know how to deal with fk.....it is changing...fk=70v..what "v"?
     
  12. Mar 20, 2007 #11
    Yes, what v? What does v mean in the equation fk = 70v and how does this relate to the equation a=(V-Vo)/t?
     
  13. Mar 20, 2007 #12
    No. That is what I just asked you. What "v". fk=70v at any given time t.
     
  14. Mar 20, 2007 #13

    Dick

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    You are NEVER going to get there this way. Throw out all of your equations that assume constant acceleration. This is ALL of them except for 'Newtons Second'. Which we will use. fk=m*a(t)=70*v(t). Now a(t)=dv(t)/dt. Put that into the preceding equation and you have a differential equation for v(t) that you need to solve. Can you handle that?
     
  15. Mar 20, 2007 #14
    You know what, you're right. I was thinking that since a = -70vt and v = v_0 + at, then v = v_0 - 70vt which one could then solve for t but v = v_0 + at is only valid when a is constant. Silly me!
     
  16. Mar 20, 2007 #15

    Dick

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    BTW, make that m*dv/dt=-70*v(t). Insert the minus sign because you know a frictional force should cause deceleration.
     
  17. Mar 20, 2007 #16
    No. But this is good news. It explains why we were not assigned this problem. The pre-req for this course was pre-calculus2, thus no differentiation...I would still like to finish this seeing as i have put so much time into it...and know a little about differentiation. I am not sure a follow what to do with this:fk=m*a(t)=70*v(t).
     
  18. Mar 20, 2007 #17
    BTW, why is it not constant acceleration?
     
  19. Mar 20, 2007 #18
    It isn't constant acceleration because acceleration is proportional to the force, and you are given in the problem that the force is proportional to the velocity. If it is accelerating (decelerating), then the velocity is not constant, and so the force is not constant, and so the acceleration is not constant.
     
  20. Mar 20, 2007 #19
    So where do I go from here?
     
  21. Mar 21, 2007 #20

    Dick

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    You now have to solve dv(t)/dt=(-k/m)*v(t). Knowing (I hope) that the derivative of exp(k*t) is k*exp(k*t), you can solve by inspection and get v(t)=C*exp(-k*t/m), where C is the speed at time 0. So what is t when v(t)=45 km/hr?
     
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