Find the best T(t) from a differential eq. and conditions

[simplex1]

An electric vehicle of mass $m$ moves along the $0-x$ axis according to the law:

m\frac{\mathrm{d} v(t)}{\mathrm{d} t} = T(t)-\mu mg-kv(t)^{2} (ma = Thrust - Friction - Drag)

It is known that:

P(t) = T(t)v(t) (Power = Thrust * Speed)

Find the thrust $T(t)$ in such a way that the vehicle be able to reach a destination as far from origin as possible with a given amount of energy, $E$, stored in its batteries, in a time \tau < \tau_{0}.

E = \int_{0}^{\tau}P(t)dt

The quantity that has to be maximized is the distance, $d$, covered:

d = \int_{0}^{\tau}v(t)dt

$E, m, \tau_{0}, g = 9.81 m/s^2, \mu, k$, the last two being the friction and drag coefficients, are all parameters with given values.

I have no idea how to solve such a problem. Some suggestions about finding $T(t)$, in Mathcad for instance, would be welcomed.

 

[Simon Bridge]

Relating a change in energy to a distance ... that would be work-energy theorem.

 

[simplex1]

This is a tough math problem. All the necessary equations have been already written.

 

[Simon Bridge]

Presumably you can also use physics.
Anyway, you wanted to know how to start... you start by playing around with the relationships and reviewing your knowedge so far like in course notes.

 

[simplex1]

I will reformulate the problem in a pure math language:

Find $x(t), y(t)$ that maximize the quantity $\int_{0}^{\tau}x(t)dt$.

It is known that:

$\frac{\mathrm{d} x(t)}{\mathrm{d} t} = a\frac{y(t)}{x(t)}-bx(t)^{2}-c$

and

$d = \int_{0}^{\tau}y(t)dt$

where $a, b, c, d, \tau$ are given parameters available as numerical values.

 

[Simon Bridge]

... you start by playing around with the relationships and reviewing your knowledge so far like in course notes.

Unless you make some attempt at the problem, it will be difficult to help you.
Experiment with the relationships, try stuff out, gain an understanding of what's going on, something should come to you.
Not everything is a matter of knowing what to do from the beginning.

i.e. One of the keys will be to realize what you are maximizing with respect to.

 

[simplex1]

Sorry but if you do not know to solve such a problem do not answer because you discourage other people that may have the solution by discrediting me with your comments suggesting that I do not know what I want when in reality the problem is clearly formulated.

 

[Tom_K]

This is an interesting problem and I would like to see the solution.
to get thrust as a f(t) does this approach help any?
Dimensionally, Distance = [(Power x Time) x (Distance / Time)] / Power
That is: Distance = Energy (Velocity/Power)
Thrust = Power/Velocity
Distance = Energy/Thrust

Thrust = E given / ∫ V dt
Now to work that Thrust (t) into the differential equation.