1. ## Expressions

Hey in the below attachment is the answers for part
i) $
a = 2x-4
$

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

and how would i do part iii) thanks again Nath

2. Originally Posted by nath_quam
Hey in the below attachment is the answers for part
i) $
a = 2x-4
$

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

and how would i do part iii) thanks again Nath
You are given:

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

so for part (i):

$
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

3. Originally Posted by nath_quam
ii) $
x = -\frac{1}{t}\ +2
$
How does this satisfy the initial condition that at $t=0, x=0\ \?$

RonL

4. yerh i see looks like i have to try again how would i do ii and iii anyway thanks nath

5. Originally Posted by nath_quam
yerh i see looks like i have to try again how would i do ii and iii anyway thanks nath
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

6. Originally Posted by nath_quam
Hey in the below attachment is the answers for part
i) $
a = 2x-4
$

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

and how would i do part iii) thanks again Nath
iii) find the distance from O when when $v=1$.

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

$
1=(2-x)^2
$
,

or:

$
x^2-4x+3=0
$
,

which can be solved using the quadratic formula, to give the $x$s
corresponding to $v=1$. The required distance is $|x|$.

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

7. Originally Posted by CaptainBlack
iii) find the distance from O when when $v=1$.

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

$
1=(2-x)^2
$
,

or:

$
x^2-4x+3=0
$
,

which can be solved using the quadratic formula, to give the $x$s
corresponding to $v=1$. The required distance is $|x|$.

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
Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time

8. Originally Posted by nath_quam
Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time
I must say that I hadn't noticed the mixed units, but I'm sure you can sort
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

9. Originally Posted by nath_quam
Two things with this part firstly the questions states velocity is in cm/s our question is in m/s and velocity isn't speed is it but distance over time
I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".

Also note:
velocity = displacement over time
speed = distance over time

-Dan

10. Originally Posted by topsquark
I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".

-Dan
Dan therefore how would i do that question part ??

Thanks for your input Nath

11. Originally Posted by nath_quam
Dan therefore how would i do that question part ??

Thanks for your input Nath
I see nothing wrong with CaptainBlack's solution. It looks like the solution is x = 1 m and x = 3 m to me.

-Dan

12. Originally Posted by topsquark
I suspect the problem is a typo. They aren't calling the unit for speed cm/s, they are calling the variable for speed "c".

-Dan
If that is the intention it leaves v undefined.

RonL

13. Originally Posted by CaptainBlack
If that is the intention it leaves v undefined.

RonL
Well, that's why I suspect a typo. Note that v and c are next to each other on the keyboard. I suspect they meant to say "v m/s" instead of "c m/s."

-Dan

14. Originally Posted by topsquark
Well, that's why I suspect a typo. Note that v and c are next to each other on the keyboard. I suspect they meant to say "v m/s" instead of "c m/s."

-Dan
I had finaly realised that is what you mean in the car between work and
home. I will add/modify the comment to the relevant post to clarify the matter.

Thanks RonL

15. Originally Posted by CaptainBlack
You are given:

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

so for part (i):

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

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