Answer
(a) Four
(b) $distance=30~m$
(c) $distance=120~m$
Work Step by Step
$v_{av}=\frac{v_{i}+v_{f}}{2}$. In both cases, $v_{f}=0$, so $v_{av}=\frac{v_{i}}{2}$. The average velocity is directly proportional to the initial velocity. But, the time required to stop is also directly proportional to the initial velocity (see problem 39). Now, $distance=(v_{av})(\Delta t)$. So, if the initial speed is doubled, both $v_{av}$ and $\Delta t$ are doubled and, consequently, the distance is increased by a factor of four.
(b) $v_{av}=\frac{16~m/s+0}{2}=8~m/s$ and $\Delta t_{1}=3.8~s$ (problem 39).
$distance=(v_{av})(\Delta t)=(8~m/s)(3.8~s)=30~m$
(c) $v_{av}=\frac{32~m/s+0}{2}=16~m/s$ and $\Delta t_{1}=7.6~s$ (problem 39).
$distance=(v_{av})(\Delta t)=(16~m/s)(7.6~s)=120~m$
