The area where one tire contacts the ground, a region about, what, 40 square inches or less? -- supports one-quarter of the weight of the car. The weight of the car does not significantly change when it is accelerating horizontally. The weight is the result of its accelerating downward as a result of gravity.

The force is the mass of the car (say, 1000 kilos) times the acceleration (9.8 meters per second per second) divided by four. Say, roughly, 2500 newtons, distributed over my hypothetical 40 square inches yields a force of about 60 newtons per square inch of force being transmitted on the foot which is run over.

This doesn't take into account the vector caused by the rotary momentum of the wheel itself, but I think that is negligible.

A newton is about the force of a Big Mac hitting the floor after falling off a table, just for comparison.

A