Question

velocity in terms of coefficient of friction formula

Answer 1

Since you have the physics tag, here is another approach:

The work done is FFd, where d is the distance travelled. The initial energy is 12mV20 (potential energy is constant here), where m is the truck mass and V0 the initial speed. The final energy is zero. We have FF=μFN=μmg, where g is the acceleration due to gravity. This gives μmgd=12mV20, or μ=12v20dg.

Using the numbers you provided, μ=12(14.2)2(25.8)(9.81)≈0.4.

or

You need to compare the force due to the truck's deceleration and the force due to friction. You know the force due to friction is (at most) μmg, so let's find the force due to deceleration.

You know, from kinematics, that

vf=v0+atf
and
xf=x0+v0tf+12at2f
where x0,v0 indicate initial position/velocity, and xf,vf,tf final position, velocity, time.

We want to know what acceleration the truck undergoes, and we know v0=14.2, vf=0, xf=25.8 (if we take x0=0).

So, in fact, we have

0=14.2+atf
25.8=14.2tf+12at2f.
You can solve these equations for tf and a (a is what you're interested in), and the force from deceleration is ma.

(Note: there's also the formula v2f=v20+2a(xf−x0), and you could simply start here to make the problem easier. The work we did above could be used to derive this formula).