Linear motion under a variable force — common mistakes

Always state which form of acceleration you are using ($a=\frac{dv}{dt}$ or $a=v\frac{dv}{dx}$). This clarifies your method and can earn you marks even if you make a later calculation error. Also, be extremely careful with signs. Define a positive direction at the start of the problem and stick to it for all vector quantities (displacement, velocity, force).

Look at the variable in the force function. If force is a function of time $t$, use $a = \frac{dv}{dt}$. If force is a function of displacement $x$, use $a = v\frac{dv}{dx}$. If force is a function of velocity $v$, you can use either, but the question will usually guide you towards finding either $v(t)$ or $v(x)$.

Only if you can prove that the net force on the particle is constant. If the force is given as a function like $F(t) = 2t$ or $F(x) = 3/x^2$, then the force is variable, acceleration is variable, and the suvat equations are not valid.

They are directly related. The work done by the net force on a particle equals the change in its kinetic energy. When you solve the equation $F(x) = mv\frac{dv}{dx}$ by integrating to get $\int F(x)\,dx = \int mv\,dv$, you are actually using the work-energy principle. The left side is the work done, and the right side integrates to $\frac{1}{2}mv^2$, representing kinetic energy.