What i never understood :)

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lefty

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what does 9.8 meters a second a second actually mean? I know its gravity :p but like what is 'a second a second'.... i could never understand. I think im just having one of these times, where i just think of random stuff and give my brain a workout, eheh. Also trying to think back to physics days...

Force = mass x acceleration..
Weight = mass x gravity..


My relative was telling me theres a formula where you can work out the force put on you in your car when your going around a corner at a certain speed or something?
 
It's more or less just physics terminology.

Gravity is just a measure of another measure. It depends how those measures are put together as to what units they come out.

IE: Force = Mass by acceleration.

Mass measured in Kg, Acceleration is m/s^2

so force is a measure of Kg x m/s^2 or Newtons

1 Newton = 1 Kg x m/s^2.

What other forces do you want explained?
 
nah all i really want to know is whats "a second a second" mean exactly :p

like its not...
9.8m/sec per 1 second
then 19.6m/s per 2 seconds
then 29.4m/s per 3 seconds

lol i'll never get this :p
 
9.8m per second squared = measure of constant acceleration
9.8m x 9.8m = 96.04m
96.04m x 96.04m = 9258.25m to infinity, theoretical limit speed of light, 299,792Km/ second squared or 186000Miles/second squared. :idea:
 
lefty said:
nah all i really want to know is whats "a second a second" mean exactly :p

like its not...
9.8m/sec per 1 second
then 19.6m/s per 2 seconds
then 29.4m/s per 3 seconds

lol i'll never get this :p

No not like that.... it stays constant. There is no"second a second" the ^-2 is only there because of the units.

acceleration = velocity/time

velocity = m/s
time = s

m/s/s = m/s^-2

(I think thats it :oops: )

All it is, is a measure of constant acceleration. 9.8 metres per second......
If an object is falling towards the earth, it gets accelerated at a rate of 9.8m/s^-2 until drag/resistance forces equal that force exerted on the object by gravity. This is called an objects terminal velocity :wink:

80ove said:
96.04m x 96.04m = 9258.25m to infinity, theoretical limit speed of light, 299,792Km/ second squared or 186000Miles/second squared.

That is really confusing :lol:
 
First off...

V = A / T

But some how it doens't come out to 9.8 m/s^-2... well technically that's perfectly correct, except that we assume that it's travellling "down", so we reverse the axis of the equation and force 9.8 m/s^-2 to become positive.... 9.8 m/s^2. Otherwise we'd have to start with the answer and work backwards.

The drag/resistance forces are entitled FRICTION in most cases and it's a pain in the ass to work it all out.
 
If I was falling head first with some nos in me & I dropped one, it would
throw a spanner in the formula hey ? Jokin' :p
 
Ah.. nice answers everyone, but complicated. Lefty, you were kind of right, I think, "per second per second" means that every second, it gains 9.8m/s of speed, so the first second it goes 9.8 m/s, the second its 19.6 m/s, so on until it reaches terminal velocity.

lol@Aldo.. what if you were facing the sky, you might save your own life. Then again, you'd probably have to add beans to the mix, NOS isn't much good by itself..
 
what you just said is you want to drop your car from a very very long way up. Terminal velocity is the greatest velocity achieved by gravity. There is no "terminal velocity" driving along a road.
 
BobWrinkle said:
what you just said is you want to drop your car from a very very long way up. Terminal velocity is the greatest velocity achieved by gravity. There is no "terminal velocity" driving along a road.

You are infact wrong. Completely wrong.


Grab a 50 KW shitbox and go for a top speed run on the highway and tell me what speed you hit. That's TERMINAL VELOCITY.

Power "pushing" the car through the air = air pushing back... point of maximum friction where acceleration = 0.

And a car would reach terminal velocity from about 150m in the air.

Along the ground it really depends how many KW your engine is. More KW = more power to push more air out of the way = higher speeds. (assuming your gearbox is able to do the speeds)
 
Shit.

I thought about that as I was writing it. But it seemed too simple...

Your sure?

It does fit in with the theory of terminal velocity...
 

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