Abaseball player is running to second base at 5.03 m/s. when he is 4.80 m from the plate he goes into a slide. the coefficient of kinetic friction between the player and the ground is 0.180, and the coefficient of static friction is 3.14. what is his speed when he reaches the plate?

Initial velocity Vi = 5.03m/s 
Distance difference D = 4.80 m  
Kinetic Friction coefficient u = 0.18 
Static Friction coefficient Uf = 3.14 
We know g = 9.81 m/s 
Calculating th edistance at which it is stopped, d = Vi^2 / 2 ug => 
 d = 5.03^2 / 2 x 0.18 x 9.81 => d = 25.3 / 3.53 = 7.167 m
 Calculating the acceleration, a = (Vf^2 - Vi^2) / [2d] => 
 Vf is 0 as it is at max distence, a = 5.03^2 / (2 x 7.167) = 1.765 m/ s^2
 Vf = Square root of Vo^2 + (2aD) = > Square root of [0.00 + 2 x 1.765 x 4.80]
= Square root of [16.944] 
Ss his speed = 4.12 m/s

You can use vf^2 = vi^2 + ad,
 where
 vf = final velocity,
 vi = initial velocity,
 a = acceleration,
 and
 d= distance. 
 mu = force of friction /force normal, so
 force of friction = mu x force normal = mu x weight in this case 
 force of friction = ma = mu x mg substituting mg for weight
 Dividing by m gives you a = mu x g so his acceleration, which is negative because he is slowing, is
 -.18(9.8)= 1.764 m/s^2
 vf = what you are looking for, so substituting in the first formula, we get 
 vf^2 = 5.03^2 + 2(-.18)(9.8)(4.8) Take the square root of both sides, and you have it.
 vf= 2.89 m/s.

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