Answer
(a) $5.7~s$
(b) $-0.18~m/s^2$
(c) More than 2.1 m/s.
Work Step by Step
(a) $v_0=2.6~m/s$, $v=1.6~m/s$, $with~constant~acceleration$
$v_{av}=\frac{v+v_0}{2}=\frac{1.6~m/s+2.6~m/s}{2}=2.1~m/s$
$v_{av}=\frac{\Delta x}{t}$. Rearrange the equation:
$t=\frac{\Delta x}{v_{av}}=\frac{12~m}{2.1~m/s}=5.7~s$
(b) $v=v_0+at$
$1.6~m/s=2.6~m/s+a(5.7~s)$
$-1.0~m/s=(5.7~s)a$
$a=\frac{-1.0~m/s}{5.7~s}=-0.18~m/s^2$
(c) In the first 6 m, the boat was traveling with greater speed than in the last 6 m. So, it took less time in the first 6 m. If the acceleration was contant and according to $\Delta v=at$, the decrease in the speed was lower in the first 6 m than in the last 6 m. So, when the boat had coasted for 6 m, its speed was more than 2.1 m/s.
