Evil dead
Mar 1, 2007, 12:46 PM
This is dedicated to some random dude called 'Capulchin':p You inspire me to become a better Scientist - I'm 16 lololol.
Basically, this is thinking about motions logically rather than just punching in numbers into formulas. In the long run, this is more efficient and ensures you never get a mechanical question wrong again. The following also involves homegenity of equations which is essential if you need to proof your answers in a Physics test or if you are kind of stumped on a question.
Legal stuff: This is the thinking and solely the thinking of Ade. My physics teacher checked over my work and approved it, nothing more nothing less. Some instances of these notes are still in construction, although I'm sure Calpulchin will iron them out for us - he's such a great guy:p
The equations of motion
v = u + at
s = 1/2(u + v) x t
s = ut + 1/2at2
v2 = u2 + 2as
where:
v = velocity (ms-1)
u = initial velocity (ms-1)
s = displacement in a straight line (m)
t = time taken (s)
a = acceleration - the increase of speed over a certain amount of time (ms-2)
Most people usually try and get the values that they've been given, punch it in, and usually get it wrong. I've devised a straight forward way to ensure that this doesn't happen.
standard symbol equation
unit equation
calculated unit equation
re-worded and easy to understand word equation
Our first equation:
v = u + at
ms-1 = ms-1 + (ms-2 x s)
ms-1 = ms-1 + ms-1
final velocity = initial velocity + average velocity in the duration of time
You see the homogeneity of the equation? The 'at' when multiplied equals ms-1 (using your indices rules and knowledge of powers). When you add that with the 'u' which also has a unit of ms-1, it will still equal 'v' which ultimately has ms-1 as it's unit.
Our second equation:
s = 1/2(u + v) x t
ms = 1/2(ms-1 + ms-1) x s
ms = 1/2(m + m)
Total distance traveled = average velocity divided by 2
For my work, my main flaw are my revised word equations. The above could be something like ' Total distance = average velocity x speed' or whatever. It is not written in stone, the main thing you should be focusing on is the homogeneity of the equation.
Our third equation:
s = ut + (1/2 x a x t2)
m = (ms-1 x s) + 1/2(ms-2 x s2)
m = m + m
Total distance traveled = distance initially + average distance whilst accelerating
The problem with this one, is what if the object moving started from rest? Then the distance initially would be 0. If that is the case, would the average distance traveled whilst accelerating be enough to compensate the whole distance?
Our forth equation:
v2 = u2 + 2as
m2s-2 = m2s-2 + 2(m2 x s2)
m2s-2 = m2s-2 + 2(m2s-2)
erm..... final velocity squared = inital velocity squared + acceleration over distance given
And that is it! Like I said before, what is above is not written in stone and I've already identified a few problems, like what if the object starts from rest, would the word equations agree? If me, Calpuchin and others bang heads to together we can easily rule out the flaws and improve my ideas on how to understand motion.
Thank you for reading
Basically, this is thinking about motions logically rather than just punching in numbers into formulas. In the long run, this is more efficient and ensures you never get a mechanical question wrong again. The following also involves homegenity of equations which is essential if you need to proof your answers in a Physics test or if you are kind of stumped on a question.
Legal stuff: This is the thinking and solely the thinking of Ade. My physics teacher checked over my work and approved it, nothing more nothing less. Some instances of these notes are still in construction, although I'm sure Calpulchin will iron them out for us - he's such a great guy:p
The equations of motion
v = u + at
s = 1/2(u + v) x t
s = ut + 1/2at2
v2 = u2 + 2as
where:
v = velocity (ms-1)
u = initial velocity (ms-1)
s = displacement in a straight line (m)
t = time taken (s)
a = acceleration - the increase of speed over a certain amount of time (ms-2)
Most people usually try and get the values that they've been given, punch it in, and usually get it wrong. I've devised a straight forward way to ensure that this doesn't happen.
standard symbol equation
unit equation
calculated unit equation
re-worded and easy to understand word equation
Our first equation:
v = u + at
ms-1 = ms-1 + (ms-2 x s)
ms-1 = ms-1 + ms-1
final velocity = initial velocity + average velocity in the duration of time
You see the homogeneity of the equation? The 'at' when multiplied equals ms-1 (using your indices rules and knowledge of powers). When you add that with the 'u' which also has a unit of ms-1, it will still equal 'v' which ultimately has ms-1 as it's unit.
Our second equation:
s = 1/2(u + v) x t
ms = 1/2(ms-1 + ms-1) x s
ms = 1/2(m + m)
Total distance traveled = average velocity divided by 2
For my work, my main flaw are my revised word equations. The above could be something like ' Total distance = average velocity x speed' or whatever. It is not written in stone, the main thing you should be focusing on is the homogeneity of the equation.
Our third equation:
s = ut + (1/2 x a x t2)
m = (ms-1 x s) + 1/2(ms-2 x s2)
m = m + m
Total distance traveled = distance initially + average distance whilst accelerating
The problem with this one, is what if the object moving started from rest? Then the distance initially would be 0. If that is the case, would the average distance traveled whilst accelerating be enough to compensate the whole distance?
Our forth equation:
v2 = u2 + 2as
m2s-2 = m2s-2 + 2(m2 x s2)
m2s-2 = m2s-2 + 2(m2s-2)
erm..... final velocity squared = inital velocity squared + acceleration over distance given
And that is it! Like I said before, what is above is not written in stone and I've already identified a few problems, like what if the object starts from rest, would the word equations agree? If me, Calpuchin and others bang heads to together we can easily rule out the flaws and improve my ideas on how to understand motion.
Thank you for reading