He doesn't have a lead foot, it just looks that way

Let $t=0$ and $x=0$ when the car is at the position closest to the person. Suppose the car is travelling in the position $x$ direction, then the car would be at position $x$ at time $t=\frac{x}{50}$ . Light takes time to travel from the car to the observer, so to the observer the car would be at position $x$ at the apparent time $t=\frac{x}{50}+\frac{\sqrt{x^2+5^2}}{c}$ . Differentiate this equation with respect to $t$ , and we get

$1=\frac{1}{50}\dot{x}+\frac{1}{c}\frac{x}{\sqrt{x^2+5^2}}\dot{x}$

$\Rightarrow \dot{x}=(\frac{1}{50}+\frac{x}{c\sqrt{x^2+5^2}})^{-1}$

Differentiating it again we get

$0=\ddot{x}(\frac{1}{50}+\frac{x}{c\sqrt{x^2+5^2}})+25\frac{\dot{x}^2}{c}(x^2+5^2)^{-3/2}$

$\Rightarrow \ddot{x}=-\frac{25}{c}(\frac{1}{50}\sqrt{x^2+25}+\frac{x}{c})^{-3}$

Hence to find maximum acceleration (or rather, deceleration), we need to find the minimum value of $\frac{1}{50}\sqrt{x^2+25}+\frac{x}{c}=\alpha$

Rearranging the above equation and squaring both sides we can obtain

$(\frac{1}{2500}-\frac{1}{c^2})x^2+\frac{2\alpha}{c}x+\frac{1}{100}-\alpha^2=0$

And by using the fact that discriminant $\geq0$ and manipulating the equations, we can obtain

$\alpha^2\geq\frac{1}{100}-\frac{25}{c^2}$

Hence,

maximum deceleration $=\frac{25}{c}\alpha^{-3}=\frac{25}{c}(\frac{1}{100}-\frac{25}{c^2})^{-3/2}=83.3\times 10^{-6}$

Mark origin at the perpendicular from the person to the road and axis along the road in the direction of motion of the car. The person is at a distance $l$ from the road (instead of using $d$ as in question so that there is no symbol confusion when doing calculus)

Suppose the person sees the car at position ${x}'$ at time $t$ . This would correspond to the car sending him light from ${x}'$ at some previous time. At that time the distance between the car and the person would be $h=\sqrt{{x}'^2+l^2}$ . Light would have take $\Delta{t}=\dfrac{h}{c}=\dfrac{\sqrt{{x}'^2+l^2}}{c}$ time to reach the person from there. So the current position of the car would be $x={x}'+v\Delta{t}$ i.e. \begin{equation} x={x}'+\dfrac{v}{c}\sqrt{{x}'^2+l^2} \tag{1} \end{equation}

The "apparent acceleration" would be given by ${a}'=\dfrac{{{d}^{2}}{x}'}{d{{t}^{2}}}$ and "apparent velocity" would be given by ${v}'=\dfrac{d{x}'}{dt}$ . For this differentiate the equation ( $1$ ) twice.

\begin{equation} v=\frac{dx}{dt}=\frac{d}{dt}\left( {x}'+\frac{v}{c}\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}} \right) \end{equation} \begin{equation} v={v}'+\frac{v}{c}\frac{2{x}'{v}'}{2\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}} \end{equation} \begin{equation} v={v}'\left( 1+\frac{v}{c}\frac{{{x}'}}{\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}} \right) \tag{2} \end{equation}

Note that the real acceleration of car is 0 \begin{equation} a=\frac{dv}{dt}=0=\frac{d}{dt}\left( {v}'\left( 1+\frac{v}{c}\frac{{{x}'}}{\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}} \right) \right) \end{equation}

\begin{equation} 0={a}'\left( 1+\frac{v}{c}\frac{{{x}'}}{\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}} \right)+{v}'\frac{v}{c}\left( \frac{\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}\,{v}'-{x}'\frac{2{x}'{v}'}{2\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}}}{{{{{x}'}}^{2}}+{{l}^{2}}} \right) \end{equation}

\begin{equation} {a}'=-\frac{v{{{{v}'}}^{2}}{{l}^{2}}}{c\left( 1+\frac{v}{c}\frac{{{x}'}}{\sqrt{{{{{x}'}}^{2}}+{{l}^{2}}}} \right){{\left( {{{{x}'}}^{2}}+{{l}^{2}} \right)}^{\frac{3}{2}}}} \end{equation} The denominator terms increase as ${x}'$ increases so maximum magnitude of acceleration occurs at ${x}'=0$ and then $v={v}'$ from equation ( $2$ ) and \begin{equation} {a}'=\frac{v{{{{v}'}}^{2}}{{l}^{2}}}{c{{\left( {{l}^{2}} \right)}^{\frac{3}{2}}}}=\frac{v{{{{v}'}}^{2}}}{cl}=\frac{v^3}{cl}=\frac{{{50}^{3}}}{3\times {{10}^{8}}\times 5}\frac{\text{m}}{\text{s}} \end{equation}

\begin{equation} {a}'=\frac{25000}{300}\frac{\mu\text{m}}{\text{s}}=\frac{250}{3}\frac{\mu\text{m}}{\text{s}}=83.33\frac{\mu\text{m}}{\text{s}} \end{equation}

Let's call the place where our the observer stands on "A", the closest point to A on the road "O", the real position of the car at a certain given time "X" and the observed position of the car at a certain given time "Y". Let $x=OX$ , $y=OY$ , $d=OA$ , and choose our time such that at $t=0$ we have $x=0$ and so $x(t)=vt$ .

The observed position and the real position of the car is not the same due to the time lag that comes from the finite nature of the speed of light. The time lag $\Delta t$ is equal to both $\frac{x-y}{v}$ and $\frac{\sqrt{y^2+d^2}}{c}$ . With these 2 equations and $x=vt$ one can solve for y:

$y=\frac{v}{c^2-v^2} (c^2t-\sqrt{c^2v^2t^2+d^2(c^2-v^2)}) \Rightarrow \frac{d^2}{dt^2}y=\ddot{y}=a_y= -\frac{v^3c^2d^2}{(c^2v^2t^2+d^2(c^2-v^2))^{3/2}}$

The largest magnitude of the apparent acceleration is at $t=0$ : $|a_y|_{max}=|a_y(t=0)|=\frac{v^3c^2}{d(c^2-v^2)^{3/2}}=8.333 \times 10^{-5}(m/s^2)=83.33 (\mu m/s^2)$