Question

A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 34 km/h at the 78-m mark. He then maintains this speed for the next 59 meters before uniformly slowing to a final speed of 30 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur?

Answer 1

The maximum acceleration experienced by the sprinter is
`0.566m/s^2` and it is experienced in the first phase when the sprinter starts his sprint.

Step-by-step 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

`v^2=u^2+2as`

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

`a=(v^2-u^2)/(2s)=(9.4^2-0^2)/(2* 78)=0.566m/s^2`

Since after that the sprinter moves at a constant velocity thus in that phase it's acceleration is
`0m/s^2`

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

`r=(8.3^2-9.4^2)/(2* 63)=-0.154m/s^2`

Upon comparing the maximum acceleration experienced by the sprinter is
`0.566m/s^2`