Answer
(a) $2.06~m/s$
(b) $9.83~m$
Work Step by Step
(a) We know that
$v_{avg}=\frac{v_i+v_f}{2}$
We plug in the known values to obtain:
$v_{avg}=\frac{0+4.12m/s}{2}=2.06m/s$
(b) The required distance can be determined as:
$d=a_{avg}t$
We plug in the known values to obtain:
$d=(2.06m/s)(4.77s)$
$d=9.83m$
