Why is the Displacement Formula ½at2?

[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

 

[berkeman]

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

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

 

[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.

 

[Voltz]

Thanks good sirs

 

[alphy]

a = acceleration

Hi there! Great question. The reason why the formula for displacement (x) is ½at2 is because it takes into account the acceleration (a) over time (t). This means that the object is changing its velocity (v) over time, and the formula takes into account this change in velocity. When you substitute in v = at, you are only considering the velocity at a specific time, not the change in velocity over time. Similarly, when you substitute in x = vt, you are only considering the displacement at a specific time, not the change in displacement over time. The ½ in the formula is there to account for the average acceleration over time. I hope this helps clarify things! Let me know if you have any other questions.