Question on Deriving Physics Equations

[BadgerBadger92]

So I am teaching myself physics in "Physics 1 For Dummies" and I have come across a chapter dealing with deriving equations, this one being the distance through using the average velocity. I have never derived equations before, so be easy on me!

So basically it starts off with s=vt (distance equals velocity times time). We take the initial and final velocity and get the average, which is v=1/2at. When this is applied to the final equation, we get s=vt=(1/2at)t

Where did this extra t come from? I thought the equation for velocity was change in distance divided by change in time? Also, why do we add on another equal sign for it? I've never taken calculus so I may need to take that first, but I was hoping someone could explain.

 

[Wee-Lamm]

So I am teaching myself physics in "Physics 1 For Dummies" and I have come across a chapter dealing with deriving equations, this one being the distance through using the average velocity. I have never derived equations before, so be easy on me!

So basically it starts off with s=vt (distance equals velocity times time). We take the initial and final velocity and get the average, which is v=1/2at. When this is applied to the final equation, we get s=vt=(1/2at)t

Where did this extra t come from? I thought the equation for velocity was change in distance divided by change in time? Also, why do we add on another equal sign for it? I've never taken calculus so I may need to take that first, but I was hoping someone could explain.

I'll use square brackets simply to separate the terms, so you can guess which 't' you are trying to wrongly discard.

s = [v] [t]
s = [1/2 at] [t]

note, v is different than vt.