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
You are given: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
Like this: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
iii) find the distance from O when when $\displaystyle v=1$.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
I must say that I hadn't noticed the mixed units, but I'm sure you can sortOriginally 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