Answer
(a) less than $3.5m/s$
(b) $-3.1m/s$
Work Step by Step
(a) We know that $\Delta K.E\propto v^2$ $\implies Fd\propto v^2$. This relation shows that if the distance is halved then the speed will decrease by a factor of $\frac{1}{\sqrt 2}$. Hence the speed will be less than $3.5m/s$.
(b) We know that
$v_{f, new}=\sqrt{\frac{1}{2}(v_{f,old}^2+v_i^2)}$
We plug in the known values to obtain:
$v_{f, new}=\sqrt{\frac{1}{2}(12)^2+\frac{1}{2}(19)^2}=15.9m/s$
$\Delta v=15.9m/s-19m/s=-3.1m/s$
