Solving Boat's Friction Force and Speed

[davidge]

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?

 

[haruspex]

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.

 

[davidge]

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
$$v = \int \frac{d^2x}{dt^2} dt$$

 

[haruspex]

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.

 

[davidge]

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 $t$?

 

[haruspex]

a isn't constant

No, but it has a well-known relationship to v. Remember, we are looking for a differential equation.

 

[davidge]

No, but it has a well-known relationship to v. Remember, we are looking for a differential equation.

Would this be

$$\frac{d^2x}{dt^2} = - 70 \frac{dx}{dt} \\
\frac{dx}{dt} = -70x \\
\Delta V = -70 \Delta x \\
\Rightarrow \Delta t = \frac{\Delta x}{\Delta v} = - \frac{1}{70}$$

 

[haruspex]

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.

 

[davidge]

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 $$- \frac{70v_o}{m}$$

 

[haruspex]

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.