#### Answer

The speed at the finish line is 12.5 m/s

#### Work Step by Step

Let $v_f$ be the speed at the end of the acceleration period. We can find an expression for the acceleration.
$v_f = v_0+(4.0~s)~a$
$v_f = 0+(4.0~s)~a$
$a = \frac{v_f}{4.0~s}$
We can find an expression for the distance $d_1$ covered during the acceleration period.
$d_1 = \frac{1}{2}a(4.0~s)^2$
$d_1 = \frac{1}{2}(\frac{v_f}{4.0~s})(4.0~s)^2$
$d_1 = (2.0~s)~v_f$
The remaining distance $d_2 $ must be covered in 6.0 seconds. We can find the final speed $v_f$. (Note that $d_2 = 100~m - d_1$)
$d_2 = (6.0~s)~v_f$
$100~m - d_1= (6.0~s)~v_f$
$100~m - (2.0~s)~v_f= (6.0~s)~v_f$
$v_f = \frac{100~m}{8.0~s}$
$v_f = 12.5~m/s$
The speed at the end of the acceleration period is 12.5 m/s. Since the speed at the finish line is the same, the speed at the finish line is 12.5 m/s