Question

Lets say, that problems are fairly simple – something, that pre-degree theoretical physics student would solve. And student does the hardest part of the task – functional reading: parsing linguistically free form text, to get input and output variables and input variable values.

For example: a problem about kinematic equations, where there are variables {a,d,t,va,vf} and few functions that describe, how thy are dependent of each-other. So using skills acquired in playing fitting blocks where thy fit, you play with the equations to get the output variable you where looking for.

In any case, there are exactly 2 possible outputs you might want and thy are (with working example):

1) Equation for that variable

Physics[have_, find_] := Solve[Flatten[{
    d == vf * t - (a * t^2) /2, (* etc. *)
    have }], find]
Physics[True, {d}]
{{d -> (1/2)*(2*t*vf - a*t^2)}}

2) Exact or general numerical value for that variable

Physics[have_, find_] := Solve[Flatten[{
    d == vf * t - (a * t^2) /2, (* etc. *)
    have }], find]
Physics[{t == 9.7, vf == -104.98, a == -9.8}, {d}]
{{d->-557.265}}

I am not sure, that I am approaching the problem correctly.

Answer 1

I think that I would probably prefer an approach like

In[1]:= Physics[find_, have_:{}] := Solve[
  {d == vf*t - (a*t^2)/2 (* , etc *)} /. have, find]

In[2]:= Physics[d]
Out[2]= {{d -> 1/2 (-a t^2 + 2 t vf)}}

In[2]:= Physics[d, {t -> 9.7, vf -> -104.98, a -> -9.8}]
Out[2]= {{d -> -557.265}}

Where the have variables are given as a list of replacement rules.

As an aside, in these types of physics problems, a nice thing to do is define your physical constants like

N[g] = -9.8;

which produces a NValues for g. Then

N[tf] = 9.7;N[vf] = -104.98;
Physics[d, {t -> tf, vf -> vf, a -> g}]
%//N

produces

{{d->1/2 (-g tf^2+2 tf vf)}}
{{d->-557.265}}