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Basic Physics
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Okay here?s a quick crash course on Physics mechanics (and applicable calculus)
ALWAYS understand the concept first!!!
The physics concept is this:
---> is derivative
Displacement-->velocity-->acceleration
derivative of displace is velocity, derivative of velocity is acceleration
to tak the derivative of something is to find the slope or rate of change. I also like to think of it as 'how fast something is changing'.
In a physical sense, what is velocity? It's how fast you're moving from point a to point b, or how fast you're chaing your displacement.
By the same token, what is accleration? when you accerlate your car, you're changing how fast you're going. So acceleration=how fast velocity is changing.
Now does it make sense that:
derivative of displace is velocity, derivative of velocity is acceleration or
velocity is how fast you're changing displcement (location), and acceleration is how fast you're changing velocity.
Conversely, antiderivative (or integral) of something is the complete opposite of derivative. hense:
==>> is integrate
acceleration==>>velocity==>>displacement
Integration is the "summing" of a function, or finding area under a curve.
So velocity is the 'sum' of all of the acceleration and
displacement is the 'sum' of all of the velocity.
This may seem a little abstract, but just remember integration is the opposite of derivative.
in short:
derive to go from left to right, integrate to go from right to left
displacement--velocity--acceleration
Now that you have an understanding of that, here's how you would FIND the derivative (and solve problems!)
calculus crash course:
The simplest and most useful derivative technique is the power rule. Here's what you do:
POWER RULE:
Given: f(x)=x^a where a is a constant and you want to find deriative of the function, f'(x), all you do is take 'a' and multiply the x by it. then subtract 1 from a.
so you have f(x)=x^a so f'(x)=a*x^(a-1)
example:
a) f(x)=x^2 differentiate and you get f'(x)=2*x^(2-1)=2x
b) f(x)=3x^2 differnetiate and you get f'(x)=2*3*x(2-1)=6x
c) here's interesting one.... f(x)=4x+12,
one of the properties of derivatives is that you can chop the function right where the add is and derive each part, then sum them together.
so you have another way to rewrite 4x+12 is 4x^1+12x^0. remember anthing to the 0th power is 1.
Then you just do the power rule for each part separated by +.
f(x)=4x^1+12x^0
differentiate
f'(x)=4*1*x^(1-1)+12*0*x^(0-1)=4
this also shows that the derivative of a constant is always 0...justify this in your mind.
Just knowing that gives you the tool to go Left to right: displacement-->velocity-->acceleration
Now you want to go right to left:
The power rule for integration is essentailly the 'backwards' of differentiation.
Given f'(x)=x^a, f(x)=(1/(a+1))*x^(a+1)+c ................... c is constant of integration
It's easier to see in an example:
example:
a) f'(x)=x^2 so integrate: f(x)=(1/(2+1))*x^(2+1)=(1/3)x^3+c
b) f'(x)=6x so integrate: f(x)=(6/2)x^(1+1)=3x^2+c remember x=x^1
c) f'(x)=4+x integrate: f(x)=(4/(0+1))*x^(0+1)+1/(1+1)*x^(1+1)=4x+(1/2)x^2+c
the plus c term is there because you 'lose' a constant whenever you differentiate. the derivative of 4x and 4x+9999999999 is the the same. However, unless you're given an initial condition(IC), you'll never know what the c is. In the OP the IC is v(2)=16, using that bit of info, you can find c.
VERY important that you put the c down and sovle for it whenever you are given initial condition.
Now you have 2 very simple tools (you'll learn LOTS more tool in your calc class). You can go from acceleration and up (increasing power of x) and displacement and downard (decreasing power of x).
now a simple exercise if to take the power rule and start diff/int. your 'formulas' in teh OP. You'll figure out that there formulas are nothing more a statement of when then formulas apply. The golden rule of doing well in intro physics is never plug and chug into equations without understanding WHY the equations are the way it is.
CLIFFS:
PHYSICS IF FUN!!!
displacement<-->velocity<-->acceleration
where --> is derivation, and <-- is integration.
don't forget +c when you do integration, solve!
edit: flipped the direction "now you have the tool to go from left to right "
ALWAYS understand the concept first!!!
The physics concept is this:
---> is derivative
Displacement-->velocity-->acceleration
derivative of displace is velocity, derivative of velocity is acceleration
to tak the derivative of something is to find the slope or rate of change. I also like to think of it as 'how fast something is changing'.
In a physical sense, what is velocity? It's how fast you're moving from point a to point b, or how fast you're chaing your displacement.
By the same token, what is accleration? when you accerlate your car, you're changing how fast you're going. So acceleration=how fast velocity is changing.
Now does it make sense that:
derivative of displace is velocity, derivative of velocity is acceleration or
velocity is how fast you're changing displcement (location), and acceleration is how fast you're changing velocity.
Conversely, antiderivative (or integral) of something is the complete opposite of derivative. hense:
==>> is integrate
acceleration==>>velocity==>>displacement
Integration is the "summing" of a function, or finding area under a curve.
So velocity is the 'sum' of all of the acceleration and
displacement is the 'sum' of all of the velocity.
This may seem a little abstract, but just remember integration is the opposite of derivative.
in short:
derive to go from left to right, integrate to go from right to left
displacement--velocity--acceleration
Now that you have an understanding of that, here's how you would FIND the derivative (and solve problems!)
calculus crash course:
The simplest and most useful derivative technique is the power rule. Here's what you do:
POWER RULE:
Given: f(x)=x^a where a is a constant and you want to find deriative of the function, f'(x), all you do is take 'a' and multiply the x by it. then subtract 1 from a.
so you have f(x)=x^a so f'(x)=a*x^(a-1)
example:
a) f(x)=x^2 differentiate and you get f'(x)=2*x^(2-1)=2x
b) f(x)=3x^2 differnetiate and you get f'(x)=2*3*x(2-1)=6x
c) here's interesting one.... f(x)=4x+12,
one of the properties of derivatives is that you can chop the function right where the add is and derive each part, then sum them together.
so you have another way to rewrite 4x+12 is 4x^1+12x^0. remember anthing to the 0th power is 1.
Then you just do the power rule for each part separated by +.
f(x)=4x^1+12x^0
differentiate
f'(x)=4*1*x^(1-1)+12*0*x^(0-1)=4
this also shows that the derivative of a constant is always 0...justify this in your mind.
Just knowing that gives you the tool to go Left to right: displacement-->velocity-->acceleration
Now you want to go right to left:
The power rule for integration is essentailly the 'backwards' of differentiation.
Given f'(x)=x^a, f(x)=(1/(a+1))*x^(a+1)+c ................... c is constant of integration
It's easier to see in an example:
example:
a) f'(x)=x^2 so integrate: f(x)=(1/(2+1))*x^(2+1)=(1/3)x^3+c
b) f'(x)=6x so integrate: f(x)=(6/2)x^(1+1)=3x^2+c remember x=x^1
c) f'(x)=4+x integrate: f(x)=(4/(0+1))*x^(0+1)+1/(1+1)*x^(1+1)=4x+(1/2)x^2+c
the plus c term is there because you 'lose' a constant whenever you differentiate. the derivative of 4x and 4x+9999999999 is the the same. However, unless you're given an initial condition(IC), you'll never know what the c is. In the OP the IC is v(2)=16, using that bit of info, you can find c.
VERY important that you put the c down and sovle for it whenever you are given initial condition.
Now you have 2 very simple tools (you'll learn LOTS more tool in your calc class). You can go from acceleration and up (increasing power of x) and displacement and downard (decreasing power of x).
now a simple exercise if to take the power rule and start diff/int. your 'formulas' in teh OP. You'll figure out that there formulas are nothing more a statement of when then formulas apply. The golden rule of doing well in intro physics is never plug and chug into equations without understanding WHY the equations are the way it is.
CLIFFS:
PHYSICS IF FUN!!!
displacement<-->velocity<-->acceleration
where --> is derivation, and <-- is integration.
don't forget +c when you do integration, solve!
edit: flipped the direction "now you have the tool to go from left to right "
One last bit about tutoring before I go to bed. Assuming that you go to UT Austin (this looks awfully like UTHW), there are many resources for you, especially if you're an engineer.
1) UTLC (UT learning center) offers free drop-in tutoring almost all day all week. There's also free 20 hour /semester of 1 on 1 tutoring if you receive ANY kinds of financial aid (grants, loans, etc) These 1 on 1 tutors are paid tutors have been tested and trained.
2) JCL and Kinsolving engineering study tables. Another place where you can get free drop-in tutoring on weeknights. Natural science people also have their own drop=in tutoring at JCL.
3) your major's (honor) society (AIAA, Sigma gamma tau, IEEE, AICHE, ASME etc). Almost all honor societies I know have free tutoring, either drop-in (often there's so few people tat it's almost like 1 on 1). These people have been in your shoe before and can often offer you advice on all of your classes.
4) Your profs and TAs!!! I know my TA was BORED to hell during his office hours because noone ever came, they can be a great help. =).
Best of luck on your (future) homeworks and don't get discouraged =).
1) UTLC (UT learning center) offers free drop-in tutoring almost all day all week. There's also free 20 hour /semester of 1 on 1 tutoring if you receive ANY kinds of financial aid (grants, loans, etc) These 1 on 1 tutors are paid tutors have been tested and trained.
2) JCL and Kinsolving engineering study tables. Another place where you can get free drop-in tutoring on weeknights. Natural science people also have their own drop=in tutoring at JCL.
3) your major's (honor) society (AIAA, Sigma gamma tau, IEEE, AICHE, ASME etc). Almost all honor societies I know have free tutoring, either drop-in (often there's so few people tat it's almost like 1 on 1). These people have been in your shoe before and can often offer you advice on all of your classes.
4) Your profs and TAs!!! I know my TA was BORED to hell during his office hours because noone ever came, they can be a great help. =).
Best of luck on your (future) homeworks and don't get discouraged =).
No i go to Penn State University, they have also the university learning center thing + extra assistance. the hw system is from webassign.
Im a 2nd semester freshman, currently exploring computer engineer. and thanks again for the help and confidence. i feel i can actually master this...
Im a 2nd semester freshman, currently exploring computer engineer. and thanks again for the help and confidence. i feel i can actually master this...
Can someone post how to solve the problems using the equations posted:
V=Vo + at
X-Xo = Vot + 1/2at^2
v^2=Vo^2 + 2a(X-Xo)
X-Xo=1/2(Vo+v)t
X-Xo=vt-1/2at^2
took physics last year but already forgot how to solve this. jchu14's solution seems kinda hard to comprehend since apparently uses calculus.
V=Vo + at
X-Xo = Vot + 1/2at^2
v^2=Vo^2 + 2a(X-Xo)
X-Xo=1/2(Vo+v)t
X-Xo=vt-1/2at^2
took physics last year but already forgot how to solve this. jchu14's solution seems kinda hard to comprehend since apparently uses calculus.
Suspicious-Teach8788
Lifer
giancoli pwned u
Ok, ive read it and taken down the examples, seems like i can understand the derivatives part, with the power rule and that a derivative of constant=0. i understand the integrals part a little. im goin to try and learn some more from my calculus txtbook and practice more examples.
Here's this solution:
You set up the initial equation:
2.5 = 0.225Vo - (0.5)*(-9.8)*(0.225)^2
Solve for v
you should get:
Vo = 10.00
Now using the equation:
v^2=Vo^2 + 2a(Y-Yo)
you know at that at the highest point, V = 0, but a = -g, and Y-Yo = Ymax
Ymax = (Vo)^2 / 2g
So Ymax = 5.10
Distance travelled ABOVE the windows = 5.10 - 2.50 = 2.60m
Now this is taking the base of the window as being 0 m
The trick in the question, as you seen to understand, that the time given is for both up and down travel, so you only use half for one direction.
The next trick is using the correct sign for the accelerations due to gravity. When an object is travelling upwards it experiences -9.8m/s^s, NOT postive accel.
--Mark
KillerCharlie
Diamond Member
Has anyone here (namely engineers) have an introductory "calculus-based physics" where you actually USED calculus other than integrating/differentiating simple functions to find displacement/velocity/acceleration? It's really a joke... they should call it algebra-based physics.
At least in the rest of the engineering classes we got our fair share of calc, ODEs, and PDEs. CFD FTW!
At least in the rest of the engineering classes we got our fair share of calc, ODEs, and PDEs. CFD FTW!
KillerCharlie
Diamond Member
I'm aerospace too... here's one equation (inequality) I'm sure you've learned by now:
Aerospace Engineering > *
🙂
Unfortunately, these equations don't apply in problem the OP posted. These equations ONLY works when the acceleration is constant. It's very useful in problems like projectile motion where the only force exerted on the object is a constant -9.8 m/s^2 due to gravity.
When the acceleration is non-constant (function of time, velocity, displacement etc), you need calculus to solve it.
kick-ass =). Aerospace engineers are so damn cool and good at phyics to boot! 😛
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