Acar is safely negotiating an unbanked circular turn at a speed of 25 m/s. the road is dry, and the maximum static frictional force acts on the tires. suddenly a long wet patch in the road decreases the maximum static frictional force to one third of its dry-road value. if the car is to continue safely around the curve, to what speed must the driver slow the car?

The velocity must be reduced to one third to stay on the road

Explanation:

The sideways force that friction must resist comes from the centrifugal acceleration due to the turn.

fc=mv2Rfc=mv2R

the frictional force is given by

ff=μmgff=μmg where μμ is the static friction coefficient

if the car is not to skid

fc≤fffc≤ff so    

mv2R≤μmgmv2R≤μmg

v≤μgR−−−−√v≤μgR

thus vv varies as the square root of μμ

so if μμ is reduced by 9, vv must be reduced by 9–√=39=3

and thus the speed must be reduced to 26 m/s

                                                                    3

the answer is a. hope this .