A boat mass M and I fill it up with fuel mass m. When it moves with velocity υ the resistive force is F=kυ^2. 1kg of fuel gives λ Joules of energy*. How much fuel (kg) is needed to sail d meters? *Suppose no energy is lost.

Newton's 2nd Law:

`sum bb F = underbrace(F)_(" force from Engine") - underbrace(k dot x^2)_(resistance)= (dp)/(dt)`

• `(dp)/(dt) = (d(M + m) dot x)/(dt)`

`= (M + m) ddot x + dot m dot x`

`implies F = (M + m) ddot x + dot m dot x + k dot x^2`

The instantaneous Power delivered by the engine is:

`P = F * dot x = (M + m) dot x ddot x + dot m dot x^2 + k dot x^3`

If

• `1" kg fuel" equiv lambda " J"`,

then:

• `m" kg fuel" equiv m lambda " J"`,

And, bearing in mind that the boat is losing mass so `dot m lt 0`:

• `P = - dot m lambda " J/s"`

`implies - dot m lambda = (M + m) dot x ddot x + dot m dot x^2 + k dot x^3`

This is looking horrible. If we simplify by assuming that the boat travels at constant velocity `v`, then :

`- dot m lambda = dot m v^2 + k v^3`

`dot m = - (k v^3)/(v^2 + lambda )`

Integrating:

`m = m_o - t (kv^3)/(lambda +v^2)`

At constant velocity time `tau` taken for the whole journey is:

• `tau = d/v`

`m(tau) = m_o - d/v * (kv^3)/(lambda + v^2)`

`=m_o - (k dv^2)/(lambda + v^2)`

So fuel used, in Kg, is:

• `m_o - (m_o - (k dv^2)/(lambda + v^2)) = (k d v^2)/(lambda + v^2)`

Data:

• I will assume constant velocity `u`, with units `m/s`.
• The resistive force is `F=kυ^2`, with units N.
• 1 kg of fuel has `lambda " Joules"`.

To sail a distance `d` (meters), the work required to maintain velocity `u` against the resistive force `F=kυ^2` would be

`"work" = F*d = kυ^2*d`

This would be in Joules. (The units of the constant k would be `"kg"/m"` for that to work out in Joules.)

Now we need to convert the above Joules to kg of fuel. From the 3rd bullet under Data above, the conversion factor I will use is `((1 "kg of fuel")/(lambda " Joules"))`

`"fuel to sail d meters" = kυ^2*d * ((1 "kg of fuel")/(lambda " Joules"))`

I said above that `kυ^2*d` would have units of Joules. Therefore, the Joules of the expression `kυ^2*d` would cancel the Joules in the denominator above. And we have the result:

`"fuel to sail d meters" = (kυ^2*d)/lambda " kg of fuel"`

I hope this helps,
Steve