Speed of a boat

  • Thread starter davidge
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Homework Statement



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?

Homework Equations



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

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?
 

Answers and Replies

  • #2
haruspex
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f=70v
The 70 has units. The equation ought to be given as f=70v kg/s.
I integrated f to get the total force.
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.
 
  • #3
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Thanks haruspex.
What is the physical meaning of ∫f.dv? I can't think of one.
I think it's actually wrong, because it would give units $$\frac{kg}{s} \frac{m²}{s^2}$$ and this is not Newtons.
The 70 has units. The equation ought to be given as ##f = 70 v \ kg/s##.
Yes, I have forgotten to mention it in the OP post.

Write the differential equation relating velocity to acceleration.
Would this be
$$v = \int \frac{d^2x}{dt^2} dt$$
 
  • #4
haruspex
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Would this be
$$v = \int \frac{d^2x}{dt^2} dt$$
That is a general true statement. I meant the DE (not an integral equation) representing the given problem, using the expression for f.
 
  • #5
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That is a general true statement. I meant the DE representing the given problem, using the expression for f.
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##?
 
  • #6
haruspex
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a isn't constant
No, but it has a well-known relationship to v. Remember, we are looking for a differential equation.
 
  • #7
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No, but it has a well-known relationship to v. Remember, we are looking for a differential equation.
Would this be

$$\frac{d^2x}{dt^2} = - 70 \frac{dx}{dt} \\
\frac{dx}{dt} = -70x \\
\Delta V = -70 \Delta x \\
\Rightarrow \Delta t = \frac{\Delta x}{\Delta v} = - \frac{1}{70}$$
 
  • #8
haruspex
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Would this be

$$\frac{d^2x}{dt^2} = - 70 \frac{dx}{dt} \\
\frac{dx}{dt} = -70x \\
\Delta V = -70 \Delta x \\
\Rightarrow \Delta t = \frac{\Delta x}{\Delta v} = - \frac{1}{70}$$
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.
 
  • #9
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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.
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 $$- \frac{70v_o}{m}$$
 
  • #10
haruspex
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Ok. So,##\frac{d^2x}{dt^2} = -70 \frac{dx}{dt} + a_o ##
No, the constant of integration comes in as you integrate, not before.
## \frac{dx}{x} = -70dt ##
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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