A sprinter practicing for the 200‐m dash accelerates uniformly from rest at A and reaches a top speed of 40 km/h at the 60‐m mark. He then maintains this speed for the next 70 meters before uniformly slowing to a final speed of 35 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?

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

Now clearing for a:

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.

massxgravity x heght=weightxheight= pe

4900=490xh

h=10