Answer
(a) $t$ = 2 h
(b) $s$ = 74.4 mi
Work Step by Step
(a) Carol is located at $x_c$ = 2.4 mi at $t_c$ = 0 h and drives at a steady $v_c$ = 36 mph, so her displacement is
\begin{equation}
s_c = x_c + v_c (t -t_c)
\end{equation}
Ann is located at $x_a$ = 0 mi at $t_a$ = 0.50 h and drives at a steady $v_a$ = 50 mph, so her displacement is
\begin{equation}
s_a = x_a + v_a( t - t_a )
\end{equation}
Equals both equations (1) and (2)
\begin{gather*}
x_c + v_c (t -t_c)= x_a + v_a( t - t_a )\\
x_c + v_c t = x_a + v_a t - v_at_a \\
v_c t - v_a t = x_a - v_at_a - x_c\\
t = \frac{x_a -x_c- v_at_a }{ v_c-v_a} \tag{3}
\end{gather*}
Get the time $t$ by
\begin{align*}
t &= \frac{x_a -x_c- v_at_a }{ v_c-v_a}\\
&= \frac{0 \mathrm{~mi} -2.4\mathrm{~mi}- (50 \mathrm{~mph})(0.50 \mathrm{~h}) }{ 36 \mathrm{~mph}-50 \mathrm{~mph}}\\
&= \boxed{2 \mathrm{~h}}
\end{align*}
(b) The position will be
\begin{equation*}
s = x_c + v_c (t -t_c) \tag{4}
\end{equation*}
We plug the values for $x_c, v_c, t$ and $t_c$ into equation (4) to get $s$
\begin{align*}
s &= x_c + v_c (t -t_c) \\
&= 2.4\mathrm{~mi} + (36 \mathrm{~mph})(2 \mathrm{~h} -0 \mathrm{~h})\\
&= \boxed{74.4 \mathrm{~mi}}
\end{align*}
