Answer
$v_{f}=10.3~m/s$
Work Step by Step
First, let's find the average acceleration:
$a_{av}=\frac{\Delta v}{\Delta t}=\frac{4.7~m/s}{5.0~s}=0.94~m/s^{2}$
If the acceleration remains the constant, we can find the increase in speed in the additional 6.0 s rearranging the equation above:
$\Delta v=(a_{av})(\Delta t)=(0.94~m/s^{2})(6.0~s)=5.6~m/s$
But, in this additional 6.0 s, the initial speed was 4.7 m/s:
$\Delta v=v_{f}-v_{i}$
$5.6~m/s=v_{f}-4.7~m/s$
$v_{f}=10.3~m/s$
