Differential Modelling?

Question

If I let the acceleration of a particle be constant a

and a drag force magnitude F actiong on the particle then

dv/dt = a - F/m if F were constnat but

F is proportinal to v(t) so F = kV(t)

and dv/dt = a-kv(t)/m where k is the drag coefficient?

so v(t) = a-k/m*integral[v(t)dt]

v(t)= a-k/m*x(t) where x(t) is the position of the particle at time t

does thaat work

Answer 1

no the particle cannot be coefficient due to the velociety of (kV(t)/F)*(a-k/m*x(t))...this obviously disproves any resonance of arguement that the particle at time t would NEVER work

i think you shold pay more attention toward magnitude rather than time