2. A person is chasing a car stopped at a traffic light at a uniform speed of $6 \mathrm{~m}$ /s. When he is $25 \mathrm{~m}$ away from the car, the traffic light changes from red to green, and the car starts to move with a uniform acceleration of $1 \mathrm{~m} / \mathrm{s}^{2}$. Then ( ).
(A) the person can catch up with the car in $7 \mathrm{~s}$
(B) the person can catch up with the car in $10 \mathrm{~s}$
(C) the person cannot catch up with the car, and the closest distance between them is $5 \mathrm{~m}$
(D) the person cannot catch up with the car, and the closest distance between them is $7 \mathrm{~m}$
2. (D).

As shown in Figure 3, suppose the car starts moving at point $C$, while the person is at point A at this moment. After $t$ seconds, the car reaches point $D$, with the distance $C D=\frac{1}{2} a t^{2}$; the person meanwhile reaches point $B$, with the distance $A B=v t$. According to the problem, the distance between them is
$$
\begin{aligned}
S & =A C+C D-A B=25+\frac{1}{2} a t^{2}-v t \\
& =\frac{1}{2} t^{2}-6 t+25=\frac{1}{2}(t-6)^{2}+7 \geqslant 7 .
\end{aligned}
$$

It is evident that the person cannot catch up with the car. The closest distance between them occurs at the instant 6 seconds after the car starts, which is $7 \mathrm{~m}$.