#### Answer

(a) $77.8~m$
(b) The cheetah's speed is 12.5 m/s.
(c) $6.25~m/s$
(d) The first 3.11 s: $19.4~m$
The second 3.11 s: $58.4~m$

#### Work Step by Step

(a) $a_{av}=\frac{1}{2}(v_f+v_0)=\frac{1}{2}(25.0~m/s+0)=12.5~m/s$
$x=x_0+v_{av}t$.
Now, assuming that the starting position is at the origin ($x_0=0$):
$x=0+(12.5~m/s)(6.22~s)=77.8~m$
(b) The velocity, in a motion with constant acceleration, varies linearly with time. That is, starting from zero, the velocity is directly proportional to the time. But, $3.11~s=\frac{6.22~s}{2}$. Now, we can conclude that, after sprinting for 3.11 s, the cheetah's speed is $12.5 m/s=\frac{25.0~m/s}{2}$.
(c) $a_{av}=\frac{1}{2}(v_f+v_0)=\frac{1}{2}(12.5~m/s+0)=12.5~m/s=6.25~m/s$
(d) The first 3.11 s:
$x=x_0+v_{av}t$
$x=0+(6.25~m/s)(3.11~s)=19.4~m$
The second 3.11 s:
In the 6.22 the cheetah has run 77.8 m. So in the second 3.11 s:
$\Delta x=77.8~m-19.4~m=58.4~m$