Please clarify:Originally Posted by nath_quam
Which integration you are talking about
what should be 2x-4
Keep Smiiling
Malay
We obtain: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
I believe the constant is -1/2 making the answer the sameOriginally Posted by CaptainBlack
why can't this part
$\displaystyle
\frac{1}{2-x}=t+const
$
be
$\displaystyle
\frac{1}{2-x}+const=t
$