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## Main Question or Discussion Point

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, lets 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 ?

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, lets 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 ?