# Displacement formula question

1. Oct 7, 2011

### Voltz

Hey, first post (:
Was just wondering why the formula for distance is

x = ½ at2

if v = at

and x = vt
then why is it ½ when you substitute in a?

x = distance
v = velocity
t = time

2. Oct 7, 2011

### Staff: Mentor

Are you familiar with integration and differentiation in calculus? That's the easist way to explain it.

3. Oct 9, 2011

### 256bits

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.

4. Oct 10, 2011

### Voltz

Thanks good sirs