A900-kilogram compact car was stopped at a traffic light when an 1,800-kilogram luxury sedan struck it from behind. the two cars became entangled due to the collision. if the luxury sedan was traveling at a velocity of 34.0 m/s north before the collision, what is the velocity of the entangled mass after the collision? a. 11.6 kg·m/s b. 22.7 kg·m/s c. 33.8 kg·m/s d. 44.9 kg·m/s

Use conservation of linear momentum.
initial momentum of stopped car (mv = 900kg*0m/s = 0)
initial momentum of moving car (mv = 1800kg*34m/s = 61,200 kgs/s)
This must be the momentum afterward.  But the total mass after the cars is entangled is (900+1800 = 2700kg)  This mass times the final velocity, V₂,  must equal the total momentum from before the collision. That is:
61,200 = 2700V₂.  Or V₂=61,200/2700 = 22.66m/s  (B)
This is slower than the original velocity so it feels right.  But be aware that the collision required energy to crush the cars.  The final velocity of such a collision would be less than this due to the loss in energy of an inelastic collision.  This is the maximum velocity of the entangled cars, provided no energy was lost in deforming the metal of the cars.

A) acceleration. this seems to be the one that makes sense.