Boat Friction Physics Problem

  • #1
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Homework Statement


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.



Homework Equations


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

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
 
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Answers and Replies

  • #2
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You have assumed that the frictional force is constant when it is not.
 
  • #3
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The frictional force changes as the velocity of the boat slows down. Your equations seem to be ignoring this fact.
 
  • #4
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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?
 
  • #5
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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.
 
  • #6
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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.....
 
  • #7
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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.....
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?
 
  • #8
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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.
 
  • #9
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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.
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?
 
  • #10
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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"?
 
  • #11
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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"?
Yes, what v? What does v mean in the equation fk = 70v and how does this relate to the equation a=(V-Vo)/t?
 
  • #12
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No. That is what I just asked you. What "v". fk=70v at any given time t.
 
  • #13
Dick
Science Advisor
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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?
 
  • #14
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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?
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!
 
  • #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.
 
  • #16
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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).
 
  • #17
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BTW, why is it not constant acceleration?
 
  • #18
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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.
 
  • #19
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So where do I go from here?
 
  • #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?
 
  • #21
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Knowing that a(t) = -70 v(t) / m and that a(t) = d / dt v(t), then

[tex]\int_{v(t_0)}^{v(t_1)} \frac{1}{v(t)} \, dv = - \int_{t_0}^{t_1} \frac{70}{m} \, dt[/tex]

where [itex]t_0[/itex] and [itex]t_1[/itex] are defined by the problem. Solving the above should give you the answer.
 
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  • #22
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Like I said, I do not know enough about differentiation......so this is all Chinese to me. Sorry, I do not understand the notation. Thanks for trying though, I do appreciate it.:smile:
 

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