How Does Friction Affect the Deceleration and Distance Traveled by a Motorboat?

[prehisto]

Homework Statement

When engine was turned off,boat with mass m was moving with speed v_0.
The force of friction F=-$\alpha$$\nu$-$\beta$v^2.
How long it would take to drop speed of boat 3 times?
Find the distance which the boat will travel in this time?

Homework Equations

The Attempt at a Solution

I think i should try to solve differental equation in form of
m$\dot{v_0}$=-$\alpha$$\nu$-$\beta$v^2

But I really don't know what to do next,could someone,please help me with some steps or tips?

 

[prehisto]

Ok,now I am pretty sure that i have to solve DE :
$\int(\frac{dv}{ \alpha v+\beta v^2})$=$\int dt$

And integration bondaries from v=v_0 to v_0/3 ?
And for other intergral offcourse its fromt=0 to t_0

Either way, how can I find the distance?

Please help .

 

[vela]

Hint: Use
$$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}.$$ This is just the chain rule.

 

[haruspex]

Ok,now I am pretty sure that i have to solve DE :
$\int(\frac{dv}{ \alpha v+\beta v^2})$=$\int dt$

And integration bondaries from v=v_0 to v_0/3 ?
And for other integral of course its from t=0 to t_0

You dropped a minus sign (as well as a factor m). You could fix the sign by changing the v limits to being from v₀/3 to v₀.
I note that in the OP you gave the force as being $-\alpha \nu -\beta v^2$. I assume you meant $-\alpha v -\beta v^2$

 

[Nickie]

First, we can rewrite the equation as:

m(dv/dt) = -αv - βv^2

To solve this differential equation, we can use the separation of variables method. First, we can divide both sides by m to get:

dv/dt = (-α/m)v - (β/m)v^2

Next, we can separate the variables by moving all terms with v to one side and all terms with t to the other side:

dv/(-αv - βv^2) = dt/m

Now, we can integrate both sides with respect to their respective variables:

∫dv/(-αv - βv^2) = ∫dt/m

The left side can be rewritten using partial fractions:

∫dv/(-αv - βv^2) = ∫dv/(αv + βv^2) = ∫(1/(αv) + 1/(βv^2))dv

We can then integrate each term separately:

∫(1/(αv) + 1/(βv^2))dv = (1/α)ln|v| - (1/βv) + C

Where C is the constant of integration. Now, we can substitute back in our original variables:

(1/α)ln|v| - (1/βv) + C = t/m

To solve for v, we can rearrange the equation:

ln|v| = (αt + Cβ)/m

And then take the exponential of both sides:

|v| = e^((αt + Cβ)/m)

Since we are only interested in the magnitude of the velocity, we can remove the absolute value:

v = e^((αt + Cβ)/m)

To solve for the constant C, we can use the initial condition given in the problem, where at t=0, v=v_0:

v_0 = e^((Cβ)/m)

Taking the natural log of both sides and rearranging, we can solve for C:

C = mln(v_0/β)

Now, we can substitute this value back into our equation for v:

v = e^((αt + mln(v_0/β)β)/m)

To find the time it takes for the boat to drop its speed by a factor of 3, we can set v=v_0/3 and