Answer:
a = 1.02[m/s^2], from resto to 60 meters.
Explanation:
In order to solve this problem we have to analyze each of the sections separately.
First, we will analyze when the sprinter starts from the rest to the point located at 60 meters. Using the following kinematics equation.
From rest to 60 meters
[tex]v_{f}^{2}= v_{o}^{2}+2*a*x\\where:\\v_{f} = final velocity =40[km/h]\\v_{o}= initial velocity =0\\x = displacement = 60 [m]\\[/tex]
Now clearing for a:
[tex]a=\frac{v_{f}^{2}-v_{o}^{2} }{2*x} \\40[km/h]*[1000m/1km]*[1h/3600s]=11.11[m/s]\\a=\frac{11.11^{2}-0 }{2*60}\\a=1.02[m/s^2][/tex]
From 60 meters to 130 meters (60+70)
There is no acceleration as he moves at a constant velocity.
From 130 meters to 200 meters
a=\frac{v_{f}^{2}-v_{o}^{2} }{2*x} \\35[km/h]*[1000m/1km]*[1h/3600s]=9.722[m/s]\\a=\frac{9.7222^{2}-11.111^2 }{2*60}\\a=-1.47[m/s^2]
The negative sign means it's slowing down.
Therefore the biggest acceleration experiences it the moment it starts from the rest to the point located at 60 meters. The acceleration between 130 meters and 200 meters is a deceleration as its speed is decreasing.
