Answer:
The maximum acceleration experienced by the sprinter is [tex]0.566m/s^2[/tex] and it is experienced in the first phase when the sprinter starts his sprint.
Explanation:
Since the sprinter attains a final speed of 34 km/h  in  runing 200 meters while starting from rest we have
Using the third equation of kinematics we have
[tex]v^2=u^2+2as[/tex]
where
v is the final speed
u is the initial speed
a is the acceleration
s is the distance covered
Since it is given that sprinter starts from rest thus u= 0 m/s and v= 34 km/h = 9.4m/s this speed is attained at s = 78 m
Applying values in the above equation we get
[tex]a=\frac{v^2-u^2}{2s}=\frac{9.4^2-0^2}{2\times 78}=0.566m/s^2[/tex]
Since after that the sprinter moves at a constant velocity thus in that phase it's acceleration is [tex]0m/s^2[/tex]
Now since the sprinter decelerates to 30 km/h or 8.33 m/s in final 63 meters thus  the deceleration experienced is again found by third equation of kinematics as
[tex]r=\frac{8.3^2-9.4^2}{2\times 63}=-0.154m/s^2[/tex]
Upon comparing the maximum acceleration experienced by the sprinter is [tex]0.566m/s^2[/tex]
