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  • hba

A particle moves along x-axis.It starts from rest at origin O and its acceleration after t sec is given by a=3-t . Calculate the displacment from O when it is at rest.

Mathematics
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  • hba
@saifoo.khan
  • hba
Can you do it ?
find antiderivative twice for a = 3-t

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Other answers:

integrate. a=3-t v = 3t - t^2/2 + C Now set it equal to zero to get time. Then integrate again, find the disp.
  • hba
Therefore, the answer is 18. Something which appeared in the (NUST) test.
  • hba
@saifoo.khan
REALLY? O_O
  • hba
Yeah
An O'levels student who have Add math can solve this too! D: D: D:
  • hba
Hmm,good :)
  • hba
What about hyperbolic functions ? You've done that ?
give me the problem, then i can tlel.
  • hba
\[\int\limits_{}^{}e^x[\sinh^{-1} x+\frac{ 1 }{ \sqrt{1+x^2} }]dx=\]
  • hba
@saifoo.khan Like this.
INtegration by parts?
idk how to deal with sinh^-1
  • hba
So you don't know hyperbolic functions ?
use e^x(f(x) + f'(x)) rule differentiation of arcsinh(x) = 1/sqrt(...)
I guess no, then. :S
  • hba
hmm No need to worry about that it's easy,you can do that later. What about finding the area from a diag ?
We have to learn that parallelogram thing. But i can find the area/vol of a curve and stuff
  • hba
Lets see. find the Area A+B |dw:1356722864399:dw|
NOOO! Give me more difficult one.
  • hba
Ok.
  • hba
find the area A |dw:1356723204131:dw|
|dw:1356723438919:dw|
Let me grab a pen and paper.
  • hba
Hmm :)
Is calc allowed?
  • hba
No not allowed+You changed the question lol :P
Sinx graph starts from zero. lol.
  • hba
Oh okay then :/
Haha.
  • hba
Now solve it .
isk what's sin(pi/4). i can only solve with the help of calc.
  • hba
180/4
area under cos is 1/sqrt2
  • hba
Sin45=0.707
0.707-0.293 ?
  • hba
@saifoo.khan Solve it and let me know :)
0.414
  • hba
Don't do it in decimals.
dang.
1/sqrt2 - (-1+sqrt2)/sqrt2
Making: (2-sqrt2)/sqrt2
Making: \[\frac{2\sqrt2-2}{2}\]
sqrt2-1 final answer?
  • hba
\[\sqrt{2}-1 \ sq.unit\] Yeah right :D
Wheo! B)
  • hba
One more. Find the area enclosed by x=y^2 and y=x-2
umm okay. solving.
  • hba
k.
1,4
area under line = 2 squnits
trying to figure out about x=y^2
1/2 final answer?
  • hba
Let me give you options lol A)9/2 sq.units B)9/4 // C)9/7 // D)7/9 // E)4/9 //
Can't fig out.
  • hba
I also couldn't figure this out lol :P
  • hba
@hartnn Can you solve it ?
  • hba
Find the area enclosed by x=y^2 and y=x-2
this is double integration problem.. right ?
  • hba
i don't think so.
plot both curves..
|dw:1356725537575:dw|
  • hba
@hartnn Yes ?
sorry, got disconnected.
|dw:1356727367179:dw| \(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy|dw:1356727499321:dw|\)
\(\int \limits_{y_1}^{y_2}f_1(x)-f_2(x)=\int \limits_{y_1}^{y_2}(y+2)-y^2dy\)
find y2 and solve the integral
sorry, y1 and y2 limits should be |dw:1356727654740:dw|

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