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How long does it take to slow this boat down?

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Homework Statement


A 1000 kg boat is traveling at 90 km/h (25 m/s) when the its engine is shut off. The magnitude of the frictional force fk is proportional to the speed v of the boat: fk = 70v; where v is in meters per second and fk is in newtons. Find the time required for the boat to slow down to 45 km/h (12.5 m/s). The answer is suppose to be 9.9 seconds, but I have yet to figure out how to approach that answer. Feel free to help me using calculus if needed.


Homework Equations


Fnet = ma
fk = [tex]\mu[/tex]kFN = 70v


The Attempt at a Solution


I first tried to write down all forces acting on the boat prior to the engine shutting down:
FEngine - fk = Fnet
maEngine - 70v = ma1

Then I wrote down the equation for when the engine did shut down:
FEngine - fk = Fnet
m*0 - 70v = m*-a2

But now I'm stuck as to where to go next.
 
Last edited:

Answers and Replies

  • #2
gneill
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What's the mathematical definition of acceleration?
 
  • #3
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Acceleration is the change in velocity or [tex]\frac{dv}{dt}[/tex]
 
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  • #4
gneill
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Okay. At any given point after engine shutoff, what is determining the acceleration of the boat? Can you write an expression for that acceleration?
 
  • #5
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The boat's acceleration should be determined by the velocity at which the engine was shut off (25 m/s) along with the frictional force opposing the boat.

FEngine - fk = FAcceleration
ma1 - fk = ma2

The acceleration of the engine should go to zero when its turned off, but it should still have velocity, so does that mean I need to rewrite the acceleration (a1) in terms of velocity?

m*0 - fk = ma2
 
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  • #6
gneill
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You're only interested in the speed of the boat from the moment the engine is turned off, after which the only force acting is that of friction. Can you write an expression for that acceleration? a = ???
 
  • #7
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[tex]a = \frac{v-v_{0}}{t}[/tex]
 
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  • #8
gneill
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That would be the average acceleration. But acceleration is not constant. What is the function of acceleration in terms of velocity?
 
  • #9
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Uh I'm not too sure here but here is what I have:

[tex]a = \frac{dv}{dt}[/tex]
[tex]\int a dt = \int dv[/tex]
 
  • #10
gneill
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Not quite; that's very general and doesn't deal with the specifics of this problem. The problem specifies a force which is a function of velocity. Since f = m*a, then f(v) = m*a(v). So write out your expression for a using the given information. It should express the instantaneous acceleration that exists at a given velocity.
 
  • #11
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Sorry if the following isn't right, but I'm slightly confused here as I initially thought acceleration was suppose to be a function of time. Regardless, going with a(v), how do you go further because the problem does not state any other information as to describe what a(v) is equal to.

[tex]ma1(v) - 70v = ma2(v)[/tex]
[tex]ma1(25) - 70*25 = ma2(25)[/tex]
[tex]a1(25)= \frac{ma2(25) + 70*25}{m}[/tex]
 
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  • #12
gneill
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Sorry if the following isn't right, but I'm slightly confused here as I initially thought acceleration was suppose to be a function of time. Regardless, going with a(v), how do you go further because the problem does not state any other information as to describe what a(v) is equal to.

[tex]ma1(v) - 70v = ma2(v)[/tex]
[tex]ma1(25) - 70*25 = ma2(25)[/tex]
[tex]a1(25)= \frac{ma2(25) + 70*25}{m}[/tex]
Yes, acceleration is a function of time. In this case, though, it's given as a function of velocity, which is in turn a function of time. Whenever you see something like that you should be thinking "differential equation".

If you sketch the bare bones first,

[tex]m a = -70 v[/tex]

[tex]m \frac{dv}{dt} = -70 v[/tex]

This looks like a separable, first order linear differential equation. The problem statement provides the "meat" to hang on these bones in terms of the given boundary values (start and end conditions for velocity).
 
  • #13
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Yes, acceleration is a function of time. In this case, though, it's given as a function of velocity, which is in turn a function of time. Whenever you see something like that you should be thinking "differential equation".

If you sketch the bare bones first,

[tex]m a = -70 v[/tex]

[tex]m \frac{dv}{dt} = -70 v[/tex]

This looks like a separable, first order linear differential equation. The problem statement provides the "meat" to hang on these bones in terms of the given boundary values (start and end conditions for velocity).
Okay now this explanation makes much more sense, and I now understand you were referring to friction when you asked me to write the equation for acceleration.


[tex]m a = -70 v[/tex]

[tex]m \frac{dv}{dt} = -70 v[/tex]

[tex]m {dv} = -70 v {dt}[/tex]

[tex]\frac{dv}{-70v} = \frac{dt}{m}[/tex]

[tex]\int\frac{dv}{-70v} = \int\frac{dt}{m}[/tex]

[tex]\frac{ln v}{-70} = \frac{t}{m} + C[/tex]

[tex]\frac{ln (25 m/s)}{-70} = \frac{0 s}{m} + C[/tex]

[tex]-0.046 = C[/tex]

[tex]\frac{ln v}{-70} = \frac{t}{m} - 0.046[/tex]

[tex]\frac{ln (12.5 m/s)}{-70} = \frac{t}{m} - 0.046[/tex]

[tex]t = 9.9 s[/tex]

Thanks for helping me through this problem, you do not know how happy I am right now ahaha! But I have one more question, could this problem be solved without using a differential equation?
 
  • #14
gneill
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Thanks for helping me through this problem, you do not know how happy I am right now ahaha! But I have one more question, could this problem be solved without using a differential equation?
Besides taking the actual boat out and measuring it? Not that I am aware of. I would be more than happy to be proven wrong!
 
  • #15
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Alright, and again thanks for your time!
 

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