Answer
a) $8.13\;\rm m/s^2,\; 2.05\;m/s^2 ,\; 0.517\;m/s^2$
b) $x=17.148\left( e^{-0.6887t} +0.6887t-1\right)$
c) $9.9\;\rm s$
Work Step by Step
a)
To find the acceleration we need to differentiate the $v_x$ relative to $t$.
$$a_x=\dfrac{d}{dt}v_x=\dfrac{d}{dt}a(1-e^{-bt})$$
$$a_x =a\dfrac{d}{dt}(1-e^{-bt})=ab e^{-bt}$$
Plugging the constant magnitudes
$$a_x=(11.81\cdot 0.6887) e^{-0.6887t}=8.1335\;e^{-0.6887t}$$
$$\boxed{a_x=8.1335\;e^{-0.6887t}}\tag 1$$
Thus, at $t=0\;\rm s$,
$$a_x= 8.1335\;e^{-0.6887\times 0}=\color{red}{\bf 8.1335}\;\rm m/s^2$$
at $t=2\;\rm s$,
$$a_x= 8.1335\;e^{-0.6887\times 2}=\color{red}{\bf 2.052}\;\rm m/s^2$$
at $t=4\;\rm s$,
$$a_x= 8.1335\;e^{-0.6887\times 4}=\color{red}{\bf 0.5175}\;\rm m/s^2$$
b)
To find the distance, we need to take the integral of $v_x$ relative to $t$.
Thus, the left side of the given formula
$$\int v_x dt=\int \dfrac{dv_x}{dt}dt=\int_0^x dx=x$$
And hence, the right side,
$$ x=\int_0^t(a-ae^{-bt})dt=\int_0^t adt-\int_0^t ae^{-bt}$$
$$ x=at\bigg|_0^t-\left( \dfrac{ae^{-bt}}{-b}\right)\bigg|_0^t$$
$$ x=at+\left( \dfrac{ae^{-bt}}{ b}-\dfrac{a}{b}\right) $$
$$ x=at+\dfrac{a}{b}\left( e^{-bt} -1\right) $$
$$\boxed{ x= \dfrac{a}{b}\left( e^{-bt} +bt-1\right) }$$
Plugging the known;
$$\boxed{ x=17.148\left( e^{-0.6887t} +0.6887t-1\right) }$$
c) You can use the trial and error method, or you can use any software calculator like WolframAlpha.
In both cases the answer is
$$t=\color{red}{\bf 9.9 }\;\rm s$$
Noting that in WolframAlpha, the two roots of time are
$$t=9.918\;\rm s, \;-3.196\;s$$
The negative root is dismissed.
