Given that,
Radius r=20m
Speed v=36km/h
Frictional coefficient μ=0.4
Now, the road is banked with an angle
θ=tan−1(v2rg)
θ=tan−1(10020×10)
θ=tan−1(12)
tanθ=0.5
When the car travels at maximum speed so that it slips upward, μR1 acts downward
So,
R1−mgcosθ−mv21rsinθ=0.....(I)
μR1+mgsinθ−mv21rcosθ=0.....(II)
Now, after solving
v21=√rgtanθ−μ1+μtanθ
v21=√20×10(0.5−0.41+0.4×0.5)
v21=√16.66
v1=4.08m/s
v1=4.08×185km/h
v1=0.816×18
v1=14.69km/h
Hence, the value of x is 0.816