1)
![3.54 m/s^2](https://img.qammunity.org/2020/formulas/physics/college/w3m0nprpiir0qnr57ufsd0tymajh3t0l0e.png)
The average acceleration of the sprinter can be found by using the following SUVAT equation:
![v^2-u^2=2ad](https://img.qammunity.org/2020/formulas/physics/high-school/jw7t56qmbiff3sbbwkf5jokdkkf45ypqyd.png)
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
In this problem,
u = 0 (the sprinter starts from rest)
v = 11.9 m/s
d = 20.0 m
Solving for a, we find the acceleration:
![a=(v^2-u^2)/(2d)=((11.9)^2)/(2(20))=3.54 m/s^2](https://img.qammunity.org/2020/formulas/physics/college/p4zdvg8v2d6qn53ukx0tv26z7gcz20azum.png)
2) 3.36 s
We can find the time needed to reach this speed by using the SUVAT equation:
![v=u+at](https://img.qammunity.org/2020/formulas/physics/middle-school/8u69t2dm31jy4f6e8h3i9msisjzkrvuvq4.png)
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
Here we have
u = 0
v = 11.9 m/s
a = 3.54 m/s^2
Solving for t, we find the time:
![t=(v-u)/(a)=(11.9)/(3.54)=3.36 s](https://img.qammunity.org/2020/formulas/physics/college/ogt6eypehs3e622ii1px2brc74myp23hss.png)