A 55 kg baseball player slides into third base with an intial speed of 4.6 m/s. if the coefficient of kinetic friction between the player and the ground is 0.46, what is the player's acceleration?

In the x-direction, the frictional force is the lone force pro tem on the player

The formula is: Fnet = ma

f = ma

In the y-direction, N, the normal force turns up and mg, the force of gravity turns down

N = mg (since there is no speeding up in the vertical direction)

f = - uN

=> f = - umg

-umg = ma

=> a = - ug

Using the kinematics equation:

d = (vf^2 - vi^2) / 2a

d = vi^2 / 2ug

d = 4.6^2 / 2 (4.508)

d = 2.35 m/s