When driver will make a circular turn then force for centripetal acceleration will be given by friction which is Ff=mv2/d But if he applies break then according to 3rd equation of motion v2−u2=2ad Putting v=0 we get a=v2/2d and force required by friction will be Ff=mv2/2d. It means when he will apply break he will need half fricional force than taking a turn. Hence driver should apply break than taking a circular turn.