Answer
Most of the time required to stop is spent braking.
Work Step by Step
As we calculated in part (c):
$T_b = 6.1~s$
We can find $T_b$ in terms of $T_r$:
$\frac{T_b}{T_r} = \frac{6100~ms}{400~ms}$
$\frac{T_b}{T_r} = 15.25$
$T_b = 15.25~T_r$
Since $T_b$ is larger than $T_r$ by a factor of $15.25$, most of the time required to stop is spent braking.
