Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
           Wₐ = 1500 lb
           W_b = 1125 lb
 Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
              W = mg
              m = W / g
              mₐ = Wₐ / g
              m_b = W_b / g
              mₐ = 1500/32 = 46.875 slug
              m_b = 125/32 = 35,156 slug
Let's reduce to the english system
              v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
            p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
            p_f = (mₐ + m_b) v
the moment is preserved
            p₀ = p_f
            mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
            v =  
we substitute the values
            v =  
            v = 0.559 v₀ₐ - 26.40                  (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
               M = mₐ + m_b
               M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
               K₀ = ½ M v²
final point. When they stop
              K_f = 0
The work is
              W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
 Let's write Newton's second law
Axis y
            N-W = 0
            N = W
the friction force has the expression
             fr = μ N
we substitute
             -μ W x = Kf - Ko
              
             -μ W x = 0 - ½ (W / g) v²
             v² = 2 μ g x  
             v =  Ra (2 0.750 32 17.5
Ra (2 0.750 32 17.5  
             v = 28.98 ft / s