I assume they gave you mu dry and mu wet
Ac = v^2/R = v^2/30
inward force needed = m Ac = m v^2/30
max inward force provided = mu m g
so
v^2 = 30 mu g
friction between tires and pavement supplies the centripetal acceleration necessary for a car to turn. Using the coefficients of friction for rubber on concrete calculate the maximum speed at which a car can round a turn of radius 30 m (a) when the road is dry; (b) when the road is wet.
2 answers
Coefficient of friction=F/R=ma/mg=a/g
Coefficient of friction of rubber on dry concrete=0.6>>0.85
So to make it accurate
coefficient=0.75
0.75=a/10
a=7.5m/s^2
Recall that..
v^2=u^2+2as
v^2=0^2+2×7.5×30
v^2=450
v=√450
v=21.2m/s
(b)coefficient of friction of rubber on wet concrete=0.45>>>0.75
coefficient=0.6
0.6=a/10
a=6m/s^2
v^2=u^2+2as
v^2=0+2×6×30
v=√360
v=18.97m/s
Coefficient of friction of rubber on dry concrete=0.6>>0.85
So to make it accurate
coefficient=0.75
0.75=a/10
a=7.5m/s^2
Recall that..
v^2=u^2+2as
v^2=0^2+2×7.5×30
v^2=450
v=√450
v=21.2m/s
(b)coefficient of friction of rubber on wet concrete=0.45>>>0.75
coefficient=0.6
0.6=a/10
a=6m/s^2
v^2=u^2+2as
v^2=0+2×6×30
v=√360
v=18.97m/s