How do you derive the equations of motion given constant acceleration using calculus?

Answer 1

To derive the kinematic equations of motion in one dimension with a given acceration 'a(t)', one begins with the definition of acceleration: the change in velocity per unit time.

average acceleration = the change in velocity/time elapsed

Acceleration, technically instantaneous acceleration, is the average acceleration over a very small interval of the velocity/time function. Instantaneous acceleration (hereafter referred to simply as 'acceleration' or 'a') is then, by extension

a = limitt-->0(instantaneous velocity1 - instantaneous velocity2)/t

which is the definition of the derivitive of instantaneous velocity ('v') with respect to time ('t'). Thus we have:

a= dv/dt

because velocity is itself change in position ('x') we can similarly derive

v= dx/dt

and

a= d2x/dt2

By the fundamental theorem of calculus:

v= integral(a)dt +C

x=integral(v)dt +C

in order to eliminate the arbitrary constant C, we use initial conditions:

v0=v(0), a0=a(0), etc.

any function representing the motion of real quantities according to the principles of classical mechanics has the value 0 for all integrals taken from an arbitrary point b to the same point b, where b is within its domain. Thus:

v(0)= 0 +C

v0=C

and so for all of the other quantities. Thus we yield:

v= v0 + integral(a)dt

x= x0 + integral(v)dt

in the special case of constant acceleration, we can take those integrals:

integral(a)dt= at

integral(v)dt= integral(v0+at)= v0t + at2/2

so our final formulae are:

v(t)=v0+at

Δx(t)=v0t+at2/2