I really dont understand this

[mardlamock]

Hi, i know this may sound stupid but, why is there a 1/2 in this 1/2at^2?. I really dont get it, i just apply the formula but i want to understand it. Thanks!

[stupid_chris]

... could we know what formula this is?

[Greywind]

Don't ever be afraid to ask questions. That and the other equations you will readily see in physics have been derived from more-complicated equations using more advanced forms of mathematics. Look for calculus-based derivations of motion equations or something of the sort, I'm sure you'll be able to find what you're looking for. That is, assuming your (1/2)at^2 equation is the simplified form of the displacement-time equation: x=xi+vidt+(1/2)adt^2, if not, then I could be completely wrong haha.

[Fractal_UK]

Hi, i know this may sound stupid but, why is there a 1/2 in this 1/2at^2?. I really dont get it, i just apply the formula but i want to understand it. Thanks!

The reason that 1/2 is there is because you are averaging the speeds. You need to consider your speed at two times: at the start and at the end of the motion you are trying to measure. At the start, your speed is at = a*0 = 0. At the end, your speed is at = a*t. So, your average speed between 0 and t is 1/2at. You then just multiply that average speed by t again to get the distance travelled, hence 1/2at^2.

[stupid_chris]

Oh okay, the acceleration formula. It can simply be explained with calculus though.

The the speed formula v = vi+at is the derivative of the distance formula, x = xi+vit+1/2at^2, so to get the distance formula you are talking about, we need to do the integer of the speed formula.

It goes like this:

v = speed

vi = initial speed

a = acceleration

t = time

x = total movement

ÃŽâ€x = change of x

ÃŽâ€t = change of t

x = total movement

dx = instantaneous change of x/differential of x

dt = instantaneous change of t/differential of t

C = integration constant

xi = initial position

So yeah, that's where it comes from. Plain old maths.

Edited July 13, 2013 by stupid_chris
better image with more precise calculations

[Bunsen]

If you're not into calculus, there's a simpler route. You already know that traveling at a steady speed v for a time t sends you a distance of vt. If you graph speed vs. time, a constant speed for a certain span of time defines a rectangle (shaded). The distance traveled is the area of that rectangle.

If you accelerate steadily, then the graph of velocity vs. time is a slanted line, and if you start with v=0 then it describes a triangle. The area of a triangle is 1/2 base*height, and this has a base of t and a height of at. QED.

You can extend this to what happens if you start with a nonzero velocity, too. The area under that line can be divided into a rectangle (with an area of initial velocity * t) and a triangle (1/2 at² again).

Calculus becomes necessary if you want to find the areas under more complicated curves.

[mardlamock]

The reason that 1/2 is there is because you are averaging the speeds. You need to consider your speed at two times: at the start and at the end of the motion you are trying to measure. At the start, your speed is at = a*0 = 0. At the end, your speed is at = a*t. So, your average speed between 0 and t is 1/2at. You then just multiply that average speed by t again to get the distance travelled, hence 1/2at^2.

Oh i see! Thanks, now i get it.

[mardlamock]

Thanks guys i now understand it!