Let $x$ be displacement as a function of time $t$ and some other physical quantity $k$ such that
$ x = f(t,k) $
Now,
1) Will the accleration $a$ be $\frac{\partial^2 x}{\partial t^2}$ or $\frac{d^2 x}{dt^2}$ ?
2) Will both of the expressions yield the same result?
Thank You
On
Your first sentence is not correct, which makes your question deeply flawed. Displacement is NOT fundamentally a function of either time or any other physical quantity $k$. So it doesn't make sense to use partial derivatives involving any other physical quantity. By using dimensional analysis with SI units, you can see the fundamental unit of displacement is the metre which is independent of time. The only instance where you would need to use partial derivatives for displacement would be when considering its components, such as $x,y,z$ components for rectangular co-ordinates. However this does not employ any other physical quantity.
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If you consider a real motion of a material point its position will be a unique function of the time. If there is an additional parameter (for example force) the point will just perform another kind of motion with a different acceleration.
Therefore the correct answer is $$a=\frac {d^2x}{dt^2}.$$
The second partial derivative of the position over time has no physical meaning.
The essential question is whether the quantity $k$ depends on $t$ as well, or not.
If $k$ does not depend on $t$ then $${dx\over dt}:=\dot x(t)={\partial f\over\partial t}(t,k)\ ,$$ and similarly for the acceleration $${d^2 x\over dt^2}:=\ddot x(t)={\partial^2 f\over\partial t^2}(t,k)\ .$$ If in fact $k=k(t)$ then $$\dot x(t)={\partial f\over\partial t}(t,k)+{\partial f\over\partial k}(t,k)\,k'(t)\ ,$$and the formula for the acceleration $\ddot x(t)$ will be even more complicated.