A55 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

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