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Speed of a boat

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data

    A boat of mass 1000 kg is moving at 25 m/s. The friction force ##f## is proportional to the speed ##v## of the boat, ##f = 70v##. How many time will take for the boat to reduce its speed to 12.5 m/s?

    2. Relevant equations

    ##\vec{F_r} = m \vec{a_r}##

    3. The attempt at a solution

    Since ##f## is proportional to ##v## at each instant ##t##, I integrated ##f## to get the total force.
    $$f_{\text{total}} = \int_{v_o}^{v_f}-70vdv$$
    the minus sign is because that force is opposite to the movement.
    Then, I assumed that the total force equals the mass times the total acceleration. Next, I substituted the value for the total acceleration from the above expression and I used it in the equation: ##V = V_o + at## to get the total time ##t##. Is this correct?
     
  2. jcsd
  3. Apr 12, 2017 #2

    haruspex

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    The 70 has units. The equation ought to be given as f=70v kg/s.
    To state your equation in full, ∫f.dv = ∫70v.dv. What is the physical meaning of ∫f.dv? I can't think of one.

    Write the differential equation relating velocity to acceleration.
     
  4. Apr 12, 2017 #3
    Thanks haruspex.
    I think it's actually wrong, because it would give units $$\frac{kg}{s} \frac{m²}{s^2}$$ and this is not Newtons.
    Yes, I have forgotten to mention it in the OP post.

    Would this be
    $$v = \int \frac{d^2x}{dt^2} dt$$
     
  5. Apr 12, 2017 #4

    haruspex

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    That is a general true statement. I meant the DE (not an integral equation) representing the given problem, using the expression for f.
     
  6. Apr 12, 2017 #5
    Ah, ok.

    $$f = -70v = ma \\ \Rightarrow a = - \frac{70v}{m}$$

    The problem is that ##a## isn't constant, so how can we substitute it in the equation ##V = V_o + at## to solve for ##t##?
     
  7. Apr 12, 2017 #6

    haruspex

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    No, but it has a well-known relationship to v. Remember, we are looking for a differential equation.
     
  8. Apr 12, 2017 #7
    Would this be

    $$\frac{d^2x}{dt^2} = - 70 \frac{dx}{dt} \\
    \frac{dx}{dt} = -70x \\
    \Delta V = -70 \Delta x \\
    \Rightarrow \Delta t = \frac{\Delta x}{\Delta v} = - \frac{1}{70}$$
     
  9. Apr 12, 2017 #8

    haruspex

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    The first integration stage is fine, except that you should allow for a constant of integration. For the second stage you need to rearrange the equation so that dt occurs on one side and only terms involving x (and dx) occur on the other.
     
  10. Apr 12, 2017 #9
    Ok. So, $$ \frac{d^2x}{dt^2} = -70 \frac{dx}{dt} + a_o \\ \frac{dx}{x} = -70dt \\ lnx = -70t + a_ot + c \\ t = \frac{lnx - c}{(-70 +a_o)}$$ where it would remain to find ##a_o##, ##c## and ##x##... I guess ##a_o## could be taken to be equal to $$- \frac{70v_o}{m}$$
     
  11. Apr 13, 2017 #10

    haruspex

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    No, the constant of integration comes in as you integrate, not before.
    That equation is after integration, and then dividing by x. You need to include the constant as part of the integration step, before dividing by x.
     
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