Kinetic energy= 1/2 m v^2
so... 1/2(1350)(12^2)
kinetic energy = 97200Joules
We can salve the problem by using the formula:

where F is the force applied, k is the spring constant and x is the stretching of the spring.
From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

I believe the acceleration would be 5m/s
All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.
Answer:
v₃ = 5 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision
Pbeforecollision = Paftercollision
(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)
where:
m₁ = mass of the truck = 3000 [kg]
v₁ = velocity of the truck = 10 [m/s]
m₂ = mass of the car = 1000 [kg]
v₂ = velocity of the car before the collision = 0 (the car is parked)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 15 [m/s]
(3000*10) + (1000*0) = (3000*v₃) + (1000*15)
30000 = 3000*v₃ + 15000
3000*v₃ = 30000 - 15000
3000*v₃ = 15000
v₃ = 5 [m/s]