Hey in the below attachment is the answers for part

i)$\displaystyle

a = 2x-4

$

ii)$\displaystyle

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

$

and how would i do part iii) thanks again Nath

Printable View

- Jul 9th 2006, 08:27 PMnath_quamExpressions
Hey in the below attachment is the answers for part

i)$\displaystyle

a = 2x-4

$

ii)$\displaystyle

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

$

and how would i do part iii) thanks again Nath - Jul 10th 2006, 12:06 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

$\displaystyle v=(2-x)^2$,

so for part (i):

$\displaystyle

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

$

If as pointed out by nath_quam v is in cm/s but acceleration is wanted in

m/s^2, in the above equation the acceleration is in cm/s^2 and

needs to be divided by 100 to convert it into m/s^2. However as Topsquark

points out there is probably a typo in the question and what appears as

"c m/s", should read "v m/s" then the dimensions in the above are OK

(unless someone else spots an error :( ).

RonL - Jul 10th 2006, 12:15 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

RonL - Jul 10th 2006, 12:26 AMnath_quam
yerh i see looks like i have to try again how would i do ii and iii anyway thanks nath

- Jul 10th 2006, 01:52 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

$\displaystyle

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

$

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

$\displaystyle

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

$,

which may be integrated to give:

$\displaystyle

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

$

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

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

RonL - Jul 10th 2006, 01:58 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

If $\displaystyle v=1$, then as $\displaystyle v=(2-x)^2$:

$\displaystyle

1=(2-x)^2

$,

or:

$\displaystyle

x^2-4x+3=0

$,

which can be solved using the quadratic formula, to give the $\displaystyle x$s

corresponding to $\displaystyle v=1$. The required distance is $\displaystyle |x|$.

There may be an implied requirement that the corresponding time be positive,

so the times at which the partical is at these positions may need to be checked.

RonL - Jul 10th 2006, 02:33 AMnath_quamQuote:

Originally Posted by**CaptainBlack**

- Jul 10th 2006, 03:03 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

that out from what I/we have done. (this is partialy a problem with having the

statement of the problem/s in an attachment, a respondent cannot see what

they are replying to on the same page as that on which they are typing the

reply - I think in future I will have to retype the question so I can see it

while replying - or maybe just open two browser windows one for the reply

and another for the question)

The second point in this case is irrelevant, as the velocity is always positive,

and so is the same thing as the speed here since speed=|velocity| (we are

talking instantaneous speeds and velocities not average speeds and velocities).

RonL - Jul 10th 2006, 03:09 AMtopsquarkQuote:

Originally Posted by**nath_quam**

*unit*for speed cm/s, they are calling the variable for speed "c".

Also note:

velocity = displacement over time

speed = distance over time

-Dan - Jul 10th 2006, 03:13 AMnath_quamQuote:

Originally Posted by**topsquark**

Thanks for your input Nath - Jul 10th 2006, 03:17 AMtopsquarkQuote:

Originally Posted by**nath_quam**

-Dan - Jul 10th 2006, 03:18 AMCaptainBlackQuote:

Originally Posted by**topsquark**

RonL - Jul 10th 2006, 03:19 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - Jul 10th 2006, 03:31 AMCaptainBlackQuote:

Originally Posted by**topsquark**

home. I will add/modify the comment to the relevant post to clarify the matter.

Thanks RonL - Jul 16th 2006, 03:59 AMnath_quamQuote:

Originally Posted by**CaptainBlack**