Physics question about deriving a one dimensional kinematic equation?

[zeromodz]

Okay, we all know that

ΔX = ViT + 1/2AT^2

Say I drop a stone from rest and I want to find its distance at a given time knowing the acceleration. Since the initial velocity is zero and its accelerating vertically we can say.

ΔY = 1/2AT^2
ΔY = (AT^2) / 2 <------------------------- Keep in mind this equation.

Now, let's start from scratch with this.

D = VT
In this case we can say.
ΔY = VT
If the stone is being dropped we can say that.
Vf = Vi + AT
V = AT

ΔY = (AT)T
ΔY = AT^2 <---------------------------------------… Now compare this equation to the one I got earlier.

Why are we dividing by 2 ?

 

[Nabeshin]

Okay, we all know that

ΔX = ViT + 1/2AT^2

Say I drop a stone from rest and I want to find its distance at a given time knowing the acceleration. Since the initial velocity is zero and its accelerating vertically we can say.

ΔY = 1/2AT^2
ΔY = (AT^2) / 2 <------------------------- Keep in mind this equation.

Now, let's start from scratch with this.

D = VT
In this case we can say.
ΔY = VT
If the stone is being dropped we can say that.
Vf = Vi + AT
V = AT

ΔY = (AT)T
ΔY = AT^2 <---------------------------------------… Now compare this equation to the one I got earlier.

Why are we dividing by 2 ?

The D=VT equation is only valid for an AVERAGE velocity. You will note that d=vt+1/2at^2 reduces to this in a=0, since average velocity is the same as instantaneous velocity.

The kinematic equations are nice for solving problems, but you really need to realize what the assumptions behind each of them is. It helps if you derive them all at least once, so you know HOW to get the equation and exactly what was assumed for that derivation.

Cheers!