Expressions

**malaygoel** · July 16th 2006, 04:22 AM

Originally Posted by nath_quam

i spotted an error we don't want to integrate answer should be 2x - 4

Please clarify:
Which integration you are talking about
what should be 2x-4

Keep Smiiling
Malay

**nath_quam** · July 16th 2006, 04:25 AM

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

**malaygoel** · July 16th 2006, 04:41 AM

Originally Posted by nath_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

He used the expression a=dv/dt =dv/dx * dx/dt to calculate acceleration. He never used integration(he used integration only for calculating x in terms of t).

keep Smiling
Malay

**nath_quam** · July 16th 2006, 04:42 AM

Well his answer is wrong do the question

**malaygoel** · July 16th 2006, 04:55 AM

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

**topsquark** · July 16th 2006, 08:17 AM

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

**CaptainBlack** · July 16th 2006, 09:31 AM

Originally Posted by nath_quam

i spotted an error we don't want to integrate answer should be 2x - 4

We obtain:

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

By differentiation. Now we know that:

$ v=(2-x)^2 $

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

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

RonL

**nath_quam** · July 16th 2006, 04:52 PM

Originally Posted by CaptainBlack

We obtain:

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

RonL

i believe it should be to the power of 1 not 3

**CaptainBlack** · July 16th 2006, 08:09 PM

Originally Posted by nath_quam

i believe it should be to the power of 1 not 3

Then please show us your working, so we can see where I have gone wrong.

RonL

**nath_quam** · July 16th 2006, 08:21 PM

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

**malaygoel** · July 16th 2006, 08:24 PM

Originally Posted by nath_quam

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

a is not dv/dx.
a is dv/dt

Malay

**nath_quam** · July 16th 2006, 08:29 PM

ah true i'll get back to you i have to think

ummm..... isn't a = the derivative of v

**nath_quam** · July 17th 2006, 03:31 AM

Originally Posted by CaptainBlack

Like this:

$ v=\frac{dx}{dt}=(2-x)^2 $

which is a differential equation of variables seperable type, and so:

$ \int\frac{dx}{(2-x)^2}=\int dt $ ,

which may be integrated to give:

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

and as $x=0$ when $t=0,\ const=1/2$ , then rearranging:

$x=\frac{4t}{2t+1}$

RonL

I believe the constant is -1/2 making the answer the same

why can't this part

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

be

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

**topsquark** · July 17th 2006, 03:55 AM

Originally Posted by nath_quam

ah true i'll get back to you i have to think

ummm..... isn't a = the derivative of v

As malaygoel stated, a is the 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

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