S= theta R

x(t) A sin (omeg t)

Vt = omeg R

At=apha R

a(t) = -omega ^2 A sin (omega t)

Vt= omega R

Yes, although logically d = d + vt + 0.5at2 is equivalent to vt + 0.5at2 = 0

Term by term:

d = [L]

vt = [LT-1]*[T] = [L]

0.5at2 = [][LT-2][T2] = [L]

where L = Length and T = Time.

cosine @-= t / vt

S=vt-16t2 solve for v is what I will assume you mean.

first pull out the t

S=t(v-16t)

then devide by t

S/t=v-16t

Then add 16t to both sides

S/t + 16t = v

This can also be written as

(S+16t2)/t = v

sin(I)/VI = sin(T)/VT where T is Refraction signal and V is speed in medium.