1)The heavier it is the more friction on the road, the faster it stops on its own. I've driven my uncles F-350 diesel truck before, stopping power is no more of an issue in that for me than it is driving a Sebring (though the blind spots in that truck are a nightmare.)
Though I may be over-simplifying the issue but, the energy required to stop a vehicle moving at a velocity v is:
E = 1/2*m*v^2
the frictional force, Fr, between a vehicle and the road is proportional to the normal force, N, and the frictional coefficient, u, assuming no slipping, but the normal force due to a surface normal to gravity, g, is simply m*g, therefore:
Fr = u*N = u*m*g
Then if we use this force to stop the vehicle, the energy energy dissipated by the friction is over the distance, d, required to stop the vehicle is:
Fr*d = u*m*g*d
then using this force to stop the vehicle leads to:
u*m*g*d = 1/2*m*v^2 --> u*g*d = 1/2*v^2
So as we can see the mass term falls out completely, and the stopping power has only to do with the frictional coefficient between the tires and the road, regardless of the mass of the vehicle; this means that the added mass of the vehicle does not provide any added benefit in terms of turning or stopping, it simply wears the tires more quickly and requires proportionally more gas=energy to accelerate the greater mass m to the velocity v to achieve a kinetic energy 1/2*m*v^2...