Non Uniform Acceleration Linear Motion

[baranij]

1st I am having problems getting a solution for final Velocity in terms of Distance traveled with a non-zero initial velocity.

Acceleration = A / Velocity - B * Velocity^2


where A is a constant related to Hp (horsepower) of the vehicle and and B is a constant related to aerodynamic drag.
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2nd simpler problem that I am also trying to solve involves:

Acceleration = m * Velocity + b where acceleration varies linearly with velocity. m & b are constants.
...this 2nd one I can solve for time and distance but not for Vf (Final Velocity) in terms of distance traveled and Vi (Initial Velocity).

time = 1/m * ln[(b-m*Vf)/(b-m*Vi)]
distance traveled = Vf/m - b/m^2 * ln(b+m*Vf) - Vi/m - b/m^2 * ln(b+m*Vi)
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How do I solve for Vf (Final Velocity) in terms of Distance traveled?

Thank you very much for any help.
Jan

 

[Irid]

I'll try to help you out with #1. Is the acceleration constant? If not, I suppose you probably know how it varies in time, so you can take these steps:

I. Obtain the final velocity in terms of time:

v(t_{\text{final}}) = \int a(t)\,dt

II. Obtain the final distance in terms of time:

s(t_{\text{final}}) = \int v(t)\, dt

III. Invert the first result to obtain time in terms of velocity:

t_{\text{final}} = f(v_{\text{final}})

IV. Plug in this time to your equation from step II and voila!

 

[baranij]

Thank you for the reply.

a(v) = A/v-B*v^2 where a=acceleration, v=velocity

The acceleration is not constant it varies depending on the velocity. Where I am getting stuck is the integrating. The "A/v" term when integrated ends up being a "ln(v)" and I can't separate "v" out of the equation anymore.

Here are the integrating questions that I need identities for:
\int (dx/dt)^(^-^1^)dt\,= ?

\int (dx/dt)^(^2^)dt\,= ?

\int (d(dx/dt)/dx) dt\,= ?

I'm almost 10 years out of school so my integrating is a little rusty. I've been looking through books and the internet can't figure out how to integrate the above 3. Any help will be much appreciated.
Thank you.

 

[Pyrrhus]

Let's see,

is this your a = f(v)

a(v) = \frac{A}{v} - Bv^{2}or

a(v) = \frac{A}{v - Bv^{2}}?

By the way use the chain rule

a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}

 

[baranij]

Your 1st one is the problem at hand:

a(v) = \frac{A}{v} - Bv^{2}

I've used the chain rule and tried to integrate but get stuck on trying to separate out V_final in terms of V_initial and \DeltaX. So far I've always had V_final in terms of \DeltaX AND time.

The integration is killing me...already been through over 10 sheets of paper over the last few days to but keep going in circles. At this point I'm not sure there is a closed form solution to the problem...I really hope there is.

Thank you,
Jan

 

[Pyrrhus]

a(v) = \frac{A}{v} - Bv^{2}

v \frac{dv}{dx} = \frac{A}{v} - Bv^{2}

We agree on this right?

Ok, now let's see your work from here.

Hint:

\frac{dv}{dx} = g(v) h(x)

\int \frac{dv}{g(v)} = \int \frac{dx}{h(x)}

Ah and by the way, the integration for \frac{dv}{g(v)} is a simple substitution.

 

[baranij]

Its been 7 years since college so I'm a bit rusty in my Integration, but don't you get:

\int \frac{dv}{g(v)} = \int {d(x)}{h(x)}

from:

\frac{dv}{dx} = g(v) h(x)

instead of:

\int \frac{dv}{g(v)} = \int \frac{dx}{h(x)}

Now I'm a bit confused as to what h(x) does for me there. Do I substitute g(v) h(x) in and try to solve:
\int {v*} {d(x)}{h(x)} = \int \frac{\frac{A}{v}-Bv^{2}}{g(v)}

 

[Pyrrhus]

That's not what I meant. Let me use another example

Say you have this ordinary differential equation (separable like yours)

\frac{dy}{dx} = (y-1)^{2}

We can do this (Note: \frac{dy}{dx} is not a fraction. It's fraction-like behavior (in some cases like separable ODE) is attributed to the chain rule)

\int \frac{dy}{(y-1)^{2}} = \int dx

Just noticed yeah it is \int g(x)dx for\frac{dv}{dx} = g(v) h(x)

did you get is

\int_{v(0)}^{v} \frac{vdv}{\frac{A}{v} - Bv^{2}} = \int_{x(0)}^{x} dx

The integral on the left is a simple substitution and the integral on the right is elemental.

 

[baranij]

\int_{x(0)}^{x} dx = \int_{v(0)}^{v} \frac{vdv}{\frac{A}{v} - Bv^{2}}

The integral on the left is a simple substitution and the integral on the right is elemental.[/QUOTE]

\int dx = \int (\frac{{v^{2}}}{A-Bv^{3}}) dv

\int dx = \int (\frac{1}{A-Bv^{3}}) v^{2} dv

u=g(x)=(A-Bv^{3}) ...&... f(u)=\frac{1}{g(x)} ...&... d(u)=g'(x)=v^{2}

\int dx =\int f(u) du = \int \frac{1}{u} du

x |^{x^{f}}_{x^{i}} = ln(u) |^{u^{f}}_{u^{i}}


x |^{x^{f}}_{x^{i}} = ln(A-Bv^{3}) |^{v^{f}}_{v^{i}}
am I getting warm? :)

 

[Pyrrhus]

You're doing ok, except that the differentiation of A - Bv^{3} is not just v^2 what about -3B?

 

[baranij]

OK, Here goes again:

\int_{x(0)}^{x} dx = \int_{v(0)}^{v} \frac{vdv}{\frac{A}{v} - Bv^{2}}

\int dx = \int (\frac{{v^{2}}}{A-Bv^{3}}) dv

\int dx = \int (\frac{1}{A-Bv^{3}}) v^{2} dv = \frac{1}{-3B}\int (\frac{1}{A-Bv^{3}}) (-3B)v^{2} dv

u=g(x)=(A-Bv^{3}) ...&... f(u)=\frac{1}{g(x)} ...&... d(u)=g'(x)=-3Bv^{2}

\int dx =\int f(u) du = \frac{1}{-3B}\int \frac{1}{u} du

x |^{x_{f}}_{x_{i}} = \frac{1}{-3B}ln(u) |^{u_{f}}_{u_{i}}


x |^{x_{f}}_{x_{i}} = \frac{1}{-3B}ln(A-Bv^{3}) |^{v_{f}}_{v_{i}}

Sloving it I get:

v_{f} = \sqrt[3]{ \frac{Bv^{3}_{i}-A}{Be^{(3B*\Delta x)}} +\frac{A}{B} }

 

[baranij]

To solve for time based on initial and final speed:

\int_{t_{0}}^{t_{F}} dt = \int_{v_{0}}^{v_{F}} \frac{dv}{\frac{A}{v} - Bv^{2}}

\int dt = \int (\frac{{v}}{A-Bv^{3}}) dv

Now that the v on top of the fraction is not a derivative of the bottom A-Bv^{3}, can you suggest a good way to integrate this function.
Thank you very much.