Answer
There is a time between 2:00 and 2:10 when the acceleration is exactly $120~mi/h^2$
Work Step by Step
Let $t_1 = 0~~$ at 2:00 pm
Let $t_2 = \frac{1}{6}~~$ at 2:10 pm
Note that 10 minutes is equal to $\frac{1}{6}~hours$
Let $v(t)$ be the car's velocity. We can assume that $v(t)$ is continuous on the time interval $[t_1, t_2]$, and differentiable on the time interval $(t_1, t_2)$
According to the Mean Value Theorem, there is a number $c$ in the interval $(t_1,t_2)$ such that $v'(c) = \frac{v(t_2)-v(t_1)}{t_2-t_1}$
Then:
$v'(c) = \frac{v(t_2)-v(t_1)}{t_2-t_1}$
$v'(c) = \frac{50~mi/h-30~mi/h}{\frac{1}{6}-0}$
$v'(c) = \frac{20~mi/h}{\frac{1}{6}~h}$
$v'(c) = 120~mi/h^2$
Since $v'(t)$ is the car's acceleration, there is a time between 2:00 and 2:10 when the acceleration is exactly $120~mi/h^2$
