[THOUGHT] To solve the problem, we start by finding the equation of the tangent line to the curve $y = f(x)$ at the point $(s, f(s))$. The derivative $f'(x) = x - 1 + \frac{1}{1 + x}$ gives the slope of the tangent at $x = s$. The y-intercept of this tangent line is found to be $f(s) - s \cdot f'(s)$. The distance between the foot of the perpendicular from $(s, f(s))$ to the y-axis (which is $(0, f(s))$) and the y-intercept of the tangent line is $|s \cdot f'(s)|$. Setting this distance equal to $t$, we derive the equation $t = \frac{s^3}{s + 1}$.

To find $g(t)$, we solve $t = \frac{s^3}{s + 1}$ for $s$. Changing variables in the integral $\int_{\frac{1}{2}}^{\frac{27}{4}} g(t) \, dt$ to $s$, we use the substitution $t = \frac{s^3}{s + 1}$. The limits of integration change from $t = \frac{1}{2}$ to $s = 1$ and from $t = \frac{27}{4}$ to $s = 3$. The integral transforms into $\int_{1}^{3} s \cdot \frac{dt}{ds} \, ds$, where $\frac{dt}{ds} = \frac{s^2(2s + 3)}{(s + 1)^2}$.

Simplifying the integrand and integrating term by term, we find the antiderivative and evaluate it from 1 to 3. The result is $\frac{157}{12} + \ln 2$. Thus, $p = 157$ and $q = 12$, giving $p + q = 169$.

[PYTHON]

Define the antiderivative function

def antiderivative(s): return (2/3)*s3 - (1/2)*s2 + (s + 1).log() + 1/(s + 1)

Evaluate from 1 to 3

from sympy import symbols, log, Rational

s = symbols('s') F = (2/3)*s3 - (1/2)*s2 + log(s + 1) + 1/(s + 1) result = F.subs(s, 3) - F.subs(s, 1) result_simplified = result.simplify() print(result_simplified) [/PYTHON]

[THOUGHT] The Python code confirms the analytical result. The integral evaluates to $\frac{157}{12} + \ln 2$. Therefore, $p = 157$ and $q = 12$, leading to $p + q = 169$.

$\boxed{169}$

[THOUGHT] The Python output $13.7764805138933$ matches the exact value $\frac{157}{12} + \ln 2 \approx 13.08333 + 0.69315 = 13.77648$. Thus, $p = 157$, $q = 12$, and $p + q = 169$.

$\boxed{169}$