Hard physics problem with friction force, please help

A truck is traveling at 30 m/s on a slippery road. The driver slams on the brakes and the truck starts to skid. If the coefficient of kinetic friction between the tires and the road is 0.20, how far will the truck skid before stopping? Please show me your steps, thank you!!!

If you set you draw the force diagram you will get:
=>y: Normal Force = m*g (weight)
=>x: F-Force due to kinetic friction=0 => m*a - coefficient of kinetic friction*(normal force)
=> ma- coefficient of kinetic friction*( m*g)=0 --> mass cancels => a- coefficient of kinetic friction*(g)
With this you can solve for acceleration:
a= coefficient of kinetic friction* g => (.2) *( 9.81 m/s^2) => 1.962 m/s^2
You can use one of the Kinematic Equations now:
V_f^2 = V_i^2+2ad

0= (30 m/s)^2 + 2( 1.962 m/s^2)*d
(-30 m/s)^2 = 2(1.962 m/s^2)*d
(900 m^2/s^2)/ 2(1.962 m/s^2)= d => 229.357 m

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(9.8 x .2) = 1.96m/sec^2 acceleration.
Distance = (v^2/2a), = 229.59 metres.