= Solution
With <Lorentz factor> $\gamma=(1-v^2/c^2)^{-1/2}$, the relativistic <momentum> and <four-momentum> are
$$
p=\gamma mv,
\qquad
P^\mu=(\gamma mc,\gamma mv).
$$
The Newtonian formula $a=F/m$ fails because $p$ is not $mv$: both the magnitude and direction of $v$ affect $dp/dt$. Since
$$
\dot\gamma=\frac{\gamma^3}{c^2}v\mathbin{\cdot}a,
$$
the <relativistic force> is
$$
F=\frac{dp}{dt}
=m\gamma\left(a+\frac{\gamma^2}{c^2}(v\mathbin{\cdot}a)v\right).
$$
Taking the dot product with $v$ gives $F\mathbin{\cdot}v=m\gamma^3v\mathbin{\cdot}a$, and substitution yields the inverse relation
$$
\boxed{a=\frac{F}{m\gamma}-\frac{(F\mathbin{\cdot}v)v}{m\gamma c^2}}.
$$
This is the requested sum of a vector parallel to $F$ and one parallel to $v$.
In the constant <electric field>, $dp/dt=qE$ and the particle starts with $p(0)=0$, so $p=qEt$. The <relativistic energy-momentum relation> implies
$$
v=\frac{c^2p}{\sqrt{m^2c^4+c^2|p|^2}}
=\boxed{\frac{qEt/m}{\sqrt{1+q^2|E|^2t^2/(m^2c^2)}}}.
$$
\b[Thus the speed increases monotonically but remains <subluminal speed>[subluminal], and $v$ tends to $c$ in the direction of $qE$ as $t\to\infty$.]
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