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Moving boats with drag force

  1. Apr 8, 2015 #1
    1. The problem statement, all variables and given/known data
    a)
    A fishing boat of mass m moves through water in the +x direction. At time t = 0, it is at location x = 0 and has speed v0 at that precise moment, the boat’s captain turns off the engines and lets the boat drift. Taking the drag force into account, Fd=-Kv^2, calculate the boats velocity as a function of time.

    b)
    A sleek speedboat also passes the point x = 0 at time t = 0. Its speed at that moment is only half the fishing boat’s speed: v0/2. Now this super-sleek speedboat is designed so well that it experiences a negligible amount of drag: Fd=0 for the speed boat. Consequently, the speedboat’s captain doesn’t even
    have his engines turned on! The fishing boat passes the speedboat, but eventually, the speedboat will catch up to the fishing boat. At what position X do the two boats meet? Write down a formula for X that is solvable and involves only known parameters. (Don’t try to solve it, it doesn’t have an analytic solution.)



    2. Relevant equations
    F=ma

    3. The attempt at a solution
    For part a I set the drag force equal to ma and solved it to be v(t)=v0-m/kt. I don't know if this is right but it makes sense to me, I have no idea how to approach part b though.
     
  2. jcsd
  3. Apr 8, 2015 #2
    You were right to solve for acceleration in ma = -Kv2 and plug your result in to the equation v = v0 + at. However, you might want to do the algebra again, as I do not think that your function is correct.
     
  4. Apr 8, 2015 #3
    Also, make sure to use parentheses to denote when you have a fraction multiplied by another value. For example, 1/2x is very different from (1/2)x.
     
  5. Apr 8, 2015 #4
    Why don't you try solving this problem using relative motion? Moreover I think the problem will get a lot simpler if you could relate boat's velocity with displacement. And I think you need to recheck your expression.
     
  6. Apr 8, 2015 #5
    I think according to the question the acceleration of second boat (the sleek speedboat) is zero.
     
  7. Apr 8, 2015 #6

    collinsmark

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    Hello kitsh,

    Welcome fo Physics Forums! :smile:

    I got a different answer.

    The way you typed in your answer is kind of ambiguous but any way you interpret it, something is not quite right. If you meant,
    [tex] v(t) = v_0 -\frac{m}{k}t [/tex]
    this has problems that it implies that the acceleration is uniform, which it is not. The acceleration is not uniform in this problem.

    If you meant
    [tex] v(t) = v_0 -\frac{m}{kt} [/tex]
    This is not right either, since the velocity blows up for small t.

    The crucial check for your answer is to put your expression for velocity back into the differential equation and see if checks out.

    I'd elaborate more, but you haven't indicated your differential equation yet. And by the way, there is a differential equation involved with this problem that must be solved.
     
    Last edited: Apr 8, 2015
  8. Apr 8, 2015 #7

    collinsmark

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    That's not right either (sorry). Yes, it's true that [itex] ma = -kv^2 [/itex], but it's not true that [itex] v = v_0 +at [/itex]. The latter equation is fine for uniform acceleration, but does not apply here. In this problem the fishing boat's velocity and acceleration are both functions of time.
     
    Last edited: Apr 8, 2015
  9. Apr 8, 2015 #8

    collinsmark

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    Hint: You can solve this problem by noting that [itex] a = \frac{dv}{dt} [/itex]. Express your [itex] ma = -kv^2 [/itex], in that form. You should find that using the "separation of variables" method of solving differential equations comes in handy here. Calculus is necessary.
     
  10. Apr 8, 2015 #9
    Actually, there is a way to solve this without differential equations or advanced calculus. You can use the second fundamental theorem of calculus to find the average value of the acceleration function over the interval v - v0, which can then be substituted into the equation v = vo + at.

    Actually, that that really doesn't help with the first part of the question very much, does it?
     
    Last edited: Apr 8, 2015
  11. Apr 8, 2015 #10

    collinsmark

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    I question whether that would work.

    You see, the equation (that we've all come to love in first year, undergrad [or perhaps highschool] physics) [itex] v = v_0 + at [/itex] is itself a solution to a differential equation. It's not magic; rather it is the solution to a differential equation itself, but a simpler one. My point is that it's a different, differential equation than the one that applies here. The [itex] v = v_0 + at [/itex] equation assumes by its very nature that acceleration is uniform.

    Allow me to explain. Suppose from the start we assume that [itex] a [/itex] is constant. We start with, [itex] \frac{dv}{dt} = a [/itex]. Then we can write, [itex] dv = adt [/itex]. When integrating, since we know [itex] a [/itex] is constant, we can pull it out of the integral, [itex] \int dv = a \int dt [/itex]. After integrating we have [itex] v = at + C [/itex], applying our initial conditions we find knowing that [itex] v(0) = v_0 [/itex], meaning that [itex] C = v_0 [/itex] we end up with the equation for uniform acceleration, [itex] v = v_0 + at [/itex]

    But that whole equation relies on being able to pull the constant [itex] a [/itex] out from under the integral. It means the equation simply does not apply when acceleration is not uniform, and cannot be used when acceleration varies with time.
     
    Last edited: Apr 8, 2015
  12. Apr 8, 2015 #11

    collinsmark

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    For the problem at hand in this thread, we can qualitatively infer the requirements of an appropriate solution.
    • [itex] v [/itex] needs to equal [itex] v_0 [/itex] when [itex] t = 0[/itex].
    • As [itex] t [/itex] approaches infinity, the velocity should trail off to zero. The velocity should approach zero with increasing time, but should never quite reach zero. If the velocity turns negative (thus turning the boat around) something is wrong.
    • And of course ultimately, if you take the derivative of the velocity function, [itex] \frac{d}{dt} \left\{ v(t) \right\} [/itex] and we take the square of the velocity function, [itex] \left( v(t) \right)^2 [/itex], they must satisfy [itex] ma = -kv^2 [/itex]. With more detail that's rewritten as [itex] m \frac{d}{dt} \left\{ v(t) \right\} = -k \left( v(t) \right)^2 [/itex].
    So once a solution is solved for, one can use these sanity checks to make sure it makes sense.
     
    Last edited: Apr 8, 2015
  13. Apr 9, 2015 #12
    I cannot recall any method to solve this differential equation without integration. Cab you please site the exact method and if possible an example.
     
    Last edited: Apr 9, 2015
  14. Apr 9, 2015 #13
    The method was mentioned above - the method of separation of variables; a common technique for solving certain types of differential equations.
     
  15. Apr 9, 2015 #14

    haruspex

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    It's not particularly advanced. As collinsmark posted, your differential equation is ##m\frac{dv}{dt}=-kv^2##. Rearrange that to get all the references to v on one side and the reference to t on the other, then integrate.
     
  16. Apr 9, 2015 #15
    I think i was not very clear in putting up my concern. Sorry for that. I am in favor of integration. If you look at the post I quoted above, it is addressed to the guy who meant integration is not necessary to solve this. He was plugging it to ## v = v_o - a*t ## . I think this is wrong. So I asked him to explain himself. I guess I'm clear now.
     
    Last edited: Apr 9, 2015
  17. Apr 9, 2015 #16
    Here is how I would have approached the question
    1. Find a reference frame (if possible inertial) such that one of the boats is stationary in it.
    2. Use ## a= v*(dv/dx) ## instead of ## a= (dv/dt) ## to directly relate velocity and displacement.
     
  18. Apr 9, 2015 #17

    haruspex

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    You are addressing (b)? kitsh hasn't solved (a) yet.
    Using a= v*(dv/dx) would be a good idea if the aim were to find v as a function of x, but (a) asks for v as a function of time.
     
  19. Apr 9, 2015 #18
    In my last comment yes I was. I think I must have mentioned. Ya and for for the (a) part, all someone needs is a little knowledge if basic calculus.
     
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