How is it that in a car crash, four 8mm bolts can anchor the seat to the car? In a car crash at for example twenty metres per second. I used suvat equations and newtons second law to work out the force as as body accelerates(negatively). I estimated that the distance travelled in the crash by the body would be roughly 0.4 metres.Even using average mass of a human and car seat the force calculated was way too large to be accurate as the tensile strength of steel would be easily exceeded. I concluded that a large portion of energy is transferred by the front of the car before it affects the body. My question is how could I find an accurate but rough figure for the force in newtons acting on each individual bolt and if anybody has any data or estimates.
 Thank you
 A: It isn't uncommon for the breaking load for M8 bolts to be 1800 kg or more. That puts four at 7200 kg, enough to statically support the weight of many vehicles with passengers.
Also important is that the bolts are not the only thing transferring the force.  In any good design, much of the force will be transferred across the mating surfaces of the seat and rails to which it is bolted.
A: You suggest $0.4\ \mathrm m$ stopping distance from $20\ \mathrm{m/s}$ velocity, which with $s=u^2/(2a)$ is a deceleration of $500\ \mathrm{m/s^2}$, or more than $50g$.
My guess is that is really the maximum deceleration your body might survive.
For $80\ \mathrm{kg}$ mass that is $40\ \mathrm{kN}$ of force.
The ultimate tensile load (from http://www.amesweb.info/Screws/Metric_Bolt_Grades_Strength.aspx) for an M8 bolt of the lowest strength class is $15.7\ \mathrm{kN}$, and can rise to $47.8\ \mathrm{kN}$ for suitably chosen bolts.
As already said by Sammy it is more the seatbelt that slows you down, so the fixing points of them to the chassis are more important, but a three-point seatbelt will have at least three of them (one per fixing point), so I would say three suitable chosen M8 bolts will be on the limit, but about right.
A: You could apply a pseudo force and treat this a static problem. 
The pseudo force is the mass of the seat and passenger times the deceleration. It acts forward on the centre of mass, applying a clockwise torque (assuming the car moves to the right). The seat pivots about the forward 2 bolts, the rear 2 bolts provide a counter-clockwise torque. Balancing the torques will give you the force in each of the rear bolts.
However, this assumes that the passenger is attached rigidly to the seat. If the passenger is not wearing a seat belt, this assumption does not hold. And seat belts are usually attached to the chassis rather than the seat, providing an additional counter-clockwise torque which relieves the stress on the bolts.
