#### Answer

The speed at the finish line is 12.5 m/s

#### Work Step by Step

We can find the speed at the end of the acceleration period;
$v = (4.0~s)~a$
We can find the distance covered during the acceleration period;
$d = \frac{1}{2}a(4.0~s)^2$
$d = (8.0~s^2)~a$
The remaining distance must be covered in 6.0 seconds. We can find the acceleration:
$6.0~s = \frac{100.0~m-(8.0~s^2)~a}{(4.0~s)~a}$
$(32.0~s^2)~a = 100.0~m$
$a = 3.125~m/s^2$
We can find the speed at the finish line;
$v = (4.0~s)~a$
$v = (4.0~s)(3.125~m/s^2)$
$v = 12.5~m/s$
The speed at the finish line is 12.5 m/s.