Quantitative finite speed of propagation property for ODE (cone of dependence) Consider the following ODE initial value problem 
\begin{align*}
&\frac{d}{dt}\Phi(t,x) = \boldsymbol{F}(t,\Phi(t,x)), & t \in [0,T], \ \ x \in \mathbb{R}^N,\\
&\Phi(0,x) = x, & x \in \mathbb{R}^N.
\end{align*}
We say that $\Phi: [0,T] \times \mathbb{R}^N \to \mathbb{R}^N$ is the flow of the ODE. 
We assume that the vector field $\boldsymbol{F}:[0,T]\times \mathbb{R}^N \to \mathbb{R}^N$ is such that  that $$\frac{|\boldsymbol{F}|}{1+|x|} \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right) + L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right),$$
that is, there exist 
\begin{align*}
&\boldsymbol{F}_1 \in L^1\left([0,T]; L^1(\mathbb{R}^N) \right)\\
&\boldsymbol{F}_2 \in L^1\left([0,T]; L^\infty(\mathbb{R}^N) \right)
\end{align*}
 such that $$\frac{\boldsymbol{F}}{1+|x|} = \boldsymbol{F}_1 + \boldsymbol{F}_2.$$

If $x \in B_{R}(0)$, what is the  truncated cone with base $B_R(0)$, which we shall call $C(T)$, such that
$$\Phi(t,x) \in C(T) $$
for all $t \in [0,T]$.
 A: 
Edited according to Martin Hairer's comment: the flow $\Phi(t,0)$ can blow up in finite (and arbitrarily small) time if the
  $L^1(0,T;L^1)$ component $F_1\neq 0$ in the Minkowski sum $\frac{F}{(1+|x|)}= F_1+F_2\in L^1(0,T;L^1) + L^1(0,T;L^\infty)$. So with the OP's assumption there
  is no hope for a reasonable answer, hence from now on we simply assume
  that $$ \frac{F}{1+|x|}\in L^1(0,T;L^\infty). $$

For simplicity let me define $\beta(s):=\|F(s,\cdot)/(1+|.|)\|_{\infty}\in L^1(0,T)$ and $B_T:=\int_0^T\beta(s)ds=\|F/(1+|x|)\|_{L^1L^\infty}$.
Writing
\begin{multline*}
|\Phi(t,x)-x|\leq \int_0^t |F(s,\Phi(s,x))|ds\\
\leq \int_0^t\frac{|F(s,\Phi(s,x))|}{1+|\Phi(s,x)|}(1+|\Phi(s,x)|)ds
\leq
\int_0^t \beta(s) (1+|\Phi(s,x)|)ds
\end{multline*}
we get, with $|\Phi(s,x)|\leq |x|+|\Phi(s,x)-x|\leq R+|\Phi(s,x)-x|$,
$$
|\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds +\int_0^t\beta(s)|\Phi(s,x)-x|ds.
$$
Applying Grönwall's inequality in its integral form (and observing that $t\mapsto\int_0^t\beta(s)ds$ is continuous nondecreasing), we can conclude that
$$
|\Phi(t,x)-x|\leq (1+R)\int_0^t\beta(s)ds \exp\left(\int_0^t\beta(s)ds\right))\leq (1+R)B_T\exp(B_T)
$$
and this gives the "truncated cone" $C(T)$.
