Hey, first post (: Was just wondering why the formula for distance is x = ½ at^{2} if v = at and x = vt then why is it ½ when you substitute in a? x = distance v = velocity t = time
Are you familiar with integration and differentiation in calculus? That's the easist way to explain it.
Nice post. Very good how you defined the variables x = vt means that there is no acceleration so a = 0, and you have to plug into the whole formula and not just part of it. ( ie with a=0 and using only half an equation, x = (1/2) at^2 becomes x = 0 ) The actual forms of the equation are: x1-x0 = v0 x t with no acceleration ie a=0 Eq 0 With constant acceleration x1-x0 = ( v0 + v1)/2 x t Eq 1 where ( v0 + v1)/2 is the average velocity Also v1 = at + v0 Eq 2 sustitute equation 1 into 2 x1 - x0 = ( v0 + at + v0 )/2 x t or x1 - x0 = v0 t + (1/2) a t^2 Eq 3 the final equation of for rectinear motion with constant acceleration is (v1)^2 - (v0)^2 = 2a(x1-x0) Eq 4 Hope that helps. you have to plug into the whole formula - not just part of it.