Answer
(a) Graph
(b) $a=3.67\,\mathrm{m/s^2}$
(c) $77\,\mathrm{m}$ and $167\,\mathrm{m}$
(d) $v=36.76\,\mathrm{m/s}$
Work Step by Step
(a) Take a look at the graph.
(b) To compute the acceleration of car A, we consider the refrence frame of car B.
Then car A will appear to start accelerating from rest, $v_0=0$, for a total time of $t=7.00\,\mathrm{s}$ and covers a displacement of $\Delta x=50.0+40.0=90.0\,\mathrm{m}$.
Using the kinematic equation,
\begin{align*}
\Delta x&=v_0t+{1\over 2}at^2\\
90.0&=0+{1\over 2}a(7.00)^2\\
a&={180\over 49.0}\\
a&=3.67\,\mathrm{m/s^2}
\end{align*}
Acceleration is independent of which inertial frame the observation is made. Hence the acceleration of car A will be the same even from the ground frame: $3.67\,\mathrm{m/s^2}$.
(c) Car B travels with a constant speed throughout, hence the distance covere by it simply,
$$\Delta x=v\Delta t=11\times 7.00=77\,\mathrm{m}$$
The total distance covered by Car A can be computed from the kinematic equation:
\begin{align*}
\Delta x&= v_0t+{1\over 2}at^2\\
&=(11)(7.00)+{1\over 2} (3.67)(7.00)^2\\
&=77+89.9\\
&=167\,\mathrm{m}
\end{align*}
(d) Car accelerated throughout the passing and hence its speed will have increased. To find the speed at the end of the procedure, we use the kinematic equation:
\begin{align*}
v^2&=v_0^2+2a\Delta x\\
&=11^2+2(3.67)(167)\\
&=121+1230\\
&=1351\\
v&=\sqrt{1351}\\
v&=36.76\,\mathrm{m/s}
\end{align*}
