This is effectively strange
Are you sure that you are taking into account the current speed in the expression of f ?
I mean that you should have
I am currently calculating the numerical solution to a vehicle in motion stopping.
I have used an Euler method where the initial kinetic energy T is calculated with 1/2mv^2.
The Euler method then reduces the kinetic energy in terms of friction and wind resistance. Tn+1 = Tn + h*f(x)
where f(x) = B + Cv^2 and B, C are the constants for friction and wind.
After each iteration I have used the new kinetic energy to calculate the velocity for the next value. i.e Tn+1 = 1/2mv^2
v = sqrt((2*Tn+1)/m)
The Euler loop then continues until the Kinetic energy is zero and nh is the distance.
When plotting the results Kinetic energy against Distance I have a line. This should be a curve should it not?
Am I calculating the velocity correctly? Are there any problems with my calculations etc. Any help would be greatly appreciated
Yes, I am pretty sure. I have worked out the first three steps by hand and it matches the first three steps in the program. I'm not sure if I'm working out the velocity in the correct way for each step though.
After I reduce the Kinetic energy I then calculate the velocity based on the new kinetic energy. v = sqrt((2*T)/m) Is this correct? or is there another way to find the new velocity.
I have the results of the program if you want me to post them
For an increment of h = 1
where B and C are always set to -3454.5 and -0.2514175 respectively.
B = umg
u is the coefficient of friction.
T0 is 10ms^-1
F = h*(B+Cv^2) where v= sqrt((2*T)/m) not actually to sure which T it is calculating Tn I assume. I think this is where the problem lies I just don't know how to go about solving it.
Your results are correct
If you calculate the speed you can find the formula with exponential
But due to the fact that
we can use the approximation which leads to the linear approximation of the square of the speed
This gives a linear trend but this is only an approximation