I do not quite understand the above situation? And is it happened before the collision?

"The slope of the road heading south is S=1.45% south and the entire road slopes to the east at 2.75%. "

I did not understand the above statement,either!

The car is sent east of the impact about 18 feet, But from center to edge of pavement is 14 feet, so the car is out of the pavement?

Without consider the above parameters. From the impact distance in the north-south direction.

I try to estimate the speed right after the impact for both car:

Assume the stopping distance are all due to friction of the road (which might not be true if car was flying after the impact)

The following are very rough estimation and base on the above assumption:

acceleration for both car assume to be the same: a=-0.565*9.8=5.537 m/s^2

v=sqrt(2*a*s)=sqrt(2*9.8*0.565*s)

19 feet=5.79 m so the speed is about 8 m/s So it took 1.4 s to fully stopped after the collision.

8 feet=2.43m so the speed is about 5.2m/s So it took 0.9s to fully stopped after the collision.

Initial speed for the truck is 43mph=69km/hr=19.2m/s

Assume conservation of momentum during collision

3650*(v+5.2)=5251*(19.2+8) so v=33.9m/s = 122 km/hr= 75.9 mph (estimated speed for the car)

Because the mass for the car is smaller than the truck and the bounce off distance for the car is less than the truck, so the speed of the car is larger than the truck.

However, the above estimate are based on very simple model. I believe local police agent should have gather more data to provide a better estimation.