Answer
$t_{c2} = -1.01\times 10^{-7}~s$
Work Step by Step
According to Carman, the length of the garage is $0.379~m$
At $t_{c1} = 0$, the back of the garage is at position $x_c = 0.379~m$
Note that $x_{c2} = 30.5~m$
We can find the time $t_{c2}$:
$t_{c2} = \frac{30.5~m-0.379~m}{-(0.9980)(3.0\times 10^8~m/s)}$
$t_{c2} = -1.01\times 10^{-7}~s$
