Please clarify:Quote:

Originally Posted bynath_quam

Which integration you are talking about

what should be 2x-4

Keep Smiiling

Malay

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- Jul 16th 2006, 04:22 AMmalaygoelQuote:

Originally Posted by**nath_quam**

Which integration you are talking about

what should be 2x-4

Keep Smiiling

Malay - Jul 16th 2006, 04:25 AMnath_quam
For part i) of the question c original post, captain black has integrated to give a in terms of x we actually want to go the other way therefore answer should be 2x - 4

- Jul 16th 2006, 04:41 AMmalaygoelQuote:

Originally Posted by**nath_quam**

keep Smiling

Malay - Jul 16th 2006, 04:42 AMnath_quam
Well his answer is wrong do the question

- Jul 16th 2006, 04:55 AMmalaygoel
I don't know whether the question is wrong or not but I found that acc. is 2x -4 when v=1m/s

Keep Smiling

Malay - Jul 16th 2006, 08:17 AMtopsquark
I have just reworked the whole question. The answers given for a) and b) in the original post are incorrect. As I mentioned before, the answers CaptainBlack came up with originally are correct.

-Dan - Jul 16th 2006, 09:31 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

$\displaystyle

a=\frac{dv}{dt}=\frac{d}{dt}(2-x)^2=2(2-x)(-1)\frac{dx}{dt}\ \ \ ..(1)

$

By differentiation. Now we know that:

$\displaystyle

v=(2-x)^2

$

but $\displaystyle v=\frac{dx}{dt}$ which we may substitute back into (1) to get:

$\displaystyle

a=\frac{dv}{dt}=2(2-x)(-1)\frac{dx}{dt}=-2(2-x)^3

$.

RonL - Jul 16th 2006, 04:52 PMnath_quamQuote:

Originally Posted by**CaptainBlack**

- Jul 16th 2006, 08:09 PMCaptainBlackQuote:

Originally Posted by**nath_quam**

RonL - Jul 16th 2006, 08:21 PMnath_quamWorking
v = (2-x)^2

expanding gives

v = 4 - 4x + x^2

a= dv/dx = 2x -4

or

v = (2-x)^2

a = dv/dx = using d/dx[f(x)^n] = f'(x)n[f(x)]^(n-1)

a = ((-1)(2))(2-x)^(2-1)

a = -2(2-x)

a = 2x - 4 - Jul 16th 2006, 08:24 PMmalaygoelQuote:

Originally Posted by**nath_quam**

a is dv/dt

Malay - Jul 16th 2006, 08:29 PMnath_quam
ah true i'll get back to you i have to think

ummm..... isn't a = the derivative of v - Jul 17th 2006, 03:31 AMnath_quamJust QuicklyQuote:

Originally Posted by**CaptainBlack**

why can't this part

$\displaystyle

\frac{1}{2-x}=t+const

$

be

$\displaystyle

\frac{1}{2-x}+const=t

$ - Jul 17th 2006, 03:55 AMtopsquarkQuote:

Originally Posted by**nath_quam**

*time*derivative of v.

As to the previous post, yes the constant can be on either side of the equation, but they would be different constants...

-Dan