Derive `vecF = mvec a` where the symbols have their usual meanings.

Explanation

Newton's `II` law of motion :
Statement : "The rate of change in the linear momentum of a body is directly proportional to the impressed force and takes place in the direction of force applied.
To show that `vecF=mveca`.
Let `m` be the mass of the body . Let `vecp_(1)` be the initial linear momentum. Let `vecp_(f)` be the final linear momentum as a result of the impressed force.
By definiation `underset(Deltat to 0)(lim)(Deltavecp)/(dt)` where `Deltavecp_(1)=vecp_(f)-vecp_(i)` and `(dvecp)/(dt)` is instantaneous acceleration.
From Newton's `II` law of motion,
`(dvecp)/(dt) prop vecF`
i.e. `vecF=k(dvecp)/(dt)`, where `'k'` is a proportionality constant and `vecp=mvecv`.
i.e. `vecF=km(dvecv)/(dt)`, where `(dvecv)/(dt)=veca` (by difination)
i.e. `vecF=kmveca`
But the magnitude of force `|vecF|` is so defined that `k=1` : We denote `|vecF|=m|veca|`
Thus `F=ma` and `vecF=mveca`.