We have the initial conditions, $v(0) = u$ and $x(0) = 0$ [we're not actually told this]. The acceleration due to gravity $g$ and the coefficient of friction $\mu'$ are constants that we do not need to find. And, $a$ is the acceleration.
we have
\begin{equation} a = -g(\sin\alpha +\mu'\cos\alpha)\\\ \end{equation}
and we need to find the distance $x$, which is
\begin{equation}x = \dfrac{u^2}{2g(\sin\alpha + \mu'\cos\alpha)}\\\ \end{equation}
I really do not know how to arrive at the equation for $x$. I assumed we could integrate twice (finding the arbitrary constants inbetween each step), but it just doesn't work - unless, I'm making a mistake.
We're not told what we're differentiating in respect to and I assumed we were differentiating in respect to $t$, but $t$ doesn't appear in the equation of $x$.
This is what my attempt went like:
\begin{equation} a(t) = \dfrac{dv}{dt} = -g(\sin\alpha +\mu'\cos\alpha)\\\ \end{equation}
so,
\begin{equation}\int a \,dt = v =g(\cos\alpha-\mu'\sin\alpha) + A\\\ \end{equation}
since $v(0) = u$
\begin{equation} v = 2g(\cos\alpha-\mu'\sin\alpha) +u\\\ \end{equation}
After this I tried the same thing but I didn't get even close to same answer.
For context, it relates to a particle moving on an inclined plane, and the equation of $x$ desribes the distance ($x$) travelled before it comes to rest.
I've been trying to solve this for the past couple of days without any luck and we can't even ask for help otherwise they fail us and if they saw this i'd get permanently kicked out. Any help is appreciated :(
Edit 1: we're not actually told if the acceleration is constant or not but can I assume it is?
Edit 2: $\alpha$ is the angle of inclination of the surface
Edit 3: the block is already moving at $t=0$ by initial speed $u$
Edit 4: I know I did the integration wrong (I was very tired), it should be $v = -g(sin\alpha +\mu'\cos\alpha)t + A = -g(sin\alpha +\mu'\cos\alpha)t + u$
One has to guess, but the task is probably to compute the distance until the object comes to rest. As $v(t)=u+at$ and then $x(t)=ut+\frac12at^2$, you then only need to insert $v(t)=0\implies t=-u/a$ into the second equation to get $x=-u^2/(2a)$.