Spele T.

asked • 01/15/16

Differential Equations Model

Given the equation for sprinter velocity, dv/dt = M(u - v)/t
 
This is an ODE of 1st order and it's LINEAR, right?

2 Answers By Expert Tutors

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Spele T.

Thanks Jim, and  you were right... I am having to generate this ODE from scratch so how about this as a more fitting equation
 
dv/dt = f(t)-v/τ
 
where dv/dt is the sprinter’s acceleration (speed), f(t) is a constant 12.1m/sec2 (the initial velocity per unit of time – initial acceleration), v(t) is the velocity at time t, and τ is 0.893 (a decay constant showing the rate of physiological exhaustion during a high-intensity, short-distance sprint.)
 
Does this look like a proper way to start? If so, I'm going move to solve for velocity and then apply some initial conditions.
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01/15/16

Jim S.

tutor
Hi Spele,
 
       First can you tell me what you are trying to do? Do you have sprinter position vs time data that you are trying to fit a model to? Or are you taking a DE course and this is a homework problem? For our purpose here I'll assume thee later.
 First why do you write f(t) and then say it a constant? The notation f(t) means the f is a function of t 
It looks like a reasonable model for a sprinters speed (in the horizontal direction) with initial conditions @t=0, v(t)=0.
This equation can be solve by separation of variables. The equation can be written as dv/(f-v)=dt integrate both sides and you get v(t)=ce-t/τ+fτ apply the initial condition to solve for the constant c. 
 
Let me know if this was helpful and if you have more questions
 
Jim
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01/16/16

Neal T. answered • 01/16/16

Tutor
5.0 (2,170)

Math!

Jim S.

tutor
Hi Neal,
        If you have some raw data (position or speed vs time) I suggest you do a least squares fit and calculate the two constants. It's a non linear least squares but it would be interesting to see if they correspond to the accepted values.
      Jim
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01/16/16

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