Another Mechanics Query (1 Viewer)

bleakarcher · Jul 17, 2011

i think we are on the same side here lol. i havent realised

hscishard · Jul 17, 2011

This confusion motivates me to stop procrastinating, and finally get some study in this holidays

bleakarcher · Jul 17, 2011

i dont think u have the idea actually. u seem to think that say when a particle is dropped from a height h and reaches a distance h units from the origin its velocity is 0.
this is wrong. in resisted motion, it is referred to as final velocity. Final velocity is the velocity of the particle during the instanteous point in time it hits the ground ie the velocity of the particle just b4 it strikes 0.

khfreakau · Jul 18, 2011

bleakarcher said:
wrong, u take the starting position as the origin alwayz

lol no.

and hscishard - my teacher said this of mechanics: "the forces bit is 4u - once you get to the actual motion bits and stuff, most mechanics questions become hard 3u level questions"

So yes, I can see why you would think it can be done by 3u methods. Also, it annoys me how in many questions it's never specified whether the resistance is an acceleration or a force. Even them, there's a question in patel which says it's a force when it's actually an acceleration >_>

bleakarcher · Jul 18, 2011

when is it you dont? i alwayz do and get the desired result, unless it is specified

Drongoski · Jul 18, 2011

khfreakau said:
lol no.

and hscishard - my teacher said this of mechanics: "the forces bit is 4u - once you get to the actual motion bits and stuff, most mechanics questions become hard 3u level questions"

So yes, I can see why you would think it can be done by 3u methods. Also, it annoys me how in many questions it's never specified whether the resistance is an acceleration or a force. Even them, there's a question in patel which says it's a force when it's actually an acceleration >_>

In maths, certain terms have assumed a well-established meaning; e.g. in a typical algebra question, "find x" does not mean you go round with a magnifying glass looking for the letter "x". Similarly "resistance" has a well-known meaning. It is, like friction, an opposing force - this has the effect of reducing the prevailing force being resisted, thereby slowing it down. This resisting force equivalently contributes a negative acceleration (deceleration) to the object. But no - resistance is a force - this force contributes a negative acceleration. So Patel is correct.

bleakarcher · Jul 18, 2011

drongoski
integral[sin^(-1)x/[1+x^2]] dx from x=-1 to 1
herez the link
http://www.angelfire.com/ab7/fourunit/cranbrook2006.pdf

Drongoski · Jul 18, 2011

bleakarcher said:
drongoski
integral[sin^(-1)x/[1+x^2]] dx from x=-1 to 1
herez the link
http://www.angelfire.com/ab7/fourunit/cranbrook2006.pdf

Sorry can't do it - too hard.

Answer: zero

Trebla · Jul 18, 2011

bleakarcher said:
integral[sin^(-1)x/[1+x^2]] dx from x=-1 to 1

The answer is zero, as the integrand is an odd function...

bleakarcher · Jul 18, 2011

ahh ok.
i havent learnt much about using the properties of even and odd functions in integrate

Trebla · Jul 18, 2011

If you want a more formalised proof:

$$Let $u=-x\Rightarrow du=-dx\\\\\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx\\\\=-\displaystyle\int_1^{-1}\dfrac{\sin^{-1}(-u)}{1+u^2}\,du\\\\=\displaystyle\int_{-1}^1\dfrac{\sin^{-1}(-u)}{1+u^2}\,du\\\\=\displaystyle\int_{-1}^1\dfrac{\sin^{-1}(-x)}{1+x^2}\,dx\quad$ as $u$ is a dummy variable$\\\\=-\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx \quad$ as $\sin^{-1}x$ is an odd function so $\sin^{-1}(-x)=-\sin^{-1}x$\\\\\Rightarrow 2\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx=0\\\\\therefore\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx=0$

Drongoski · Jul 18, 2011

Over the interval (1,-1) sin^-1 x is an odd function while 1 + x² is an even function. Therefore, their quotient:

$\frac {sin^{-1}x}{1+x^2}$ is an odd function.

Bobbo1 · Jul 18, 2011

Trebla said:
If you want a more formalised proof:

$$Let $u=-x\Rightarrow du=-dx\\\\\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx\\\\=-\displaystyle\int_1^{-1}\dfrac{\sin^{-1}(-u)}{1+u^2}\,du\\\\=\displaystyle\int_{-1}^1\dfrac{\sin^{-1}(-u)}{1+u^2}\,du\\\\=\displaystyle\int_{-1}^1\dfrac{\sin^{-1}(-x)}{1+x^2}\,dx\quad$ as $u$ is a dummy variable$\\\\=-\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx \quad$ as $\sin^{-1}x$ is an odd function so $\sin^{-1}(-x)=-\sin^{-1}x$\\\\\Rightarrow 2\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx=0\\\\\therefore\displaystyle\int_{-1}^1\dfrac{\sin^{-1}x}{1+x^2}\,dx=0$

This is so beautiful!!! (and not being sarcastic)
Once you get used to the other you forget the actual theory, lol

failwhale · Jul 18, 2011

yeah u do when v=0 x=0 because it hasn't fallen height, h, as of yet.

hscishard · Jul 18, 2011

bleakarcher said:
i dont think u have the idea actually. u seem to think that say when a particle is dropped from a height h and reaches a distance h units from the origin its velocity is 0.
this is wrong. in resisted motion, it is referred to as final velocity. Final velocity is the velocity of the particle during the instanteous point in time it hits the ground ie the velocity of the particle just b4 it strikes 0.

You're correct, but I never said that it is 0. I said it wouldn't be 0. It's just that arj1 saying when x=h, v=0 confused me a bit

Forces aye... boy mechanics is going to be fun!

Hermes1 · Jul 18, 2011

hscishard said:
You're correct, but I never said that it is 0. I said it wouldn't be 0. It's just that arj1 saying when x=h, v=0 confused me a bit

Forces aye... boy mechanics is going to be fun!

lol i cant believe this thread is still going. and hscishard trust me mechanics is the best topic in 4 unit by far.

Another Mechanics Query (1 Viewer)

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