Let $$a=mx-c$$ where $m,c$ are both positive.
I want to find the acceleration time-graph so I did:
\begin{align} a & =mx-c, \\ v \frac{dv}{dx} & =mx-c , \quad \text{then:}\\ v^2 & =mx^2-cx+c_1, \\ \end{align}
but I couldn't go any further. Can you help please
Note that your differential equation is $$ \ddot x(t)=a(x(t))=mx(t)-c $$ so multiplication with $\dot x$ and integration leads to the energy level of the solution $$ E=\frac12 \dot x^2-\frac m2\left(x-\frac cm\right)^2 $$ Now you need to solve the first order differential equation $$ \dot x(t)=\pm\sqrt{E+m\left(x-\frac cm\right)^2} $$ Try the substitution $x=\frac cm+a\sinh(u)$ for $E\ne 0$. Inserting its solution into $a(x(t))$ will give you the general acceleration-time relation.