What is the greatest speed at which a car having a track width of `1.5m` can turn in a circular path of radius `24.5m` without overturning ? Assume that the center of gravity of the car is `0.49m` above the ground. Take `g = 9.8m//s^(2)`.

Answer

`19.17 ms^(-1)` .

Explanation

Here, `b = 1.5m, r =246.49 = 24.65N`
`h = 0.49 m , upsilon =` ?
As the road is unbanked , therefore, for equilibrium,
`(m upsilon)/(r) xx h = mg xx ((b)/(2))`
or `upsilon = sqrt((g r b)/(2 h)) = sqrt((9.8 xx 24 .5 xx1.5)/(2 xx 0.49))`
`=19.17m//s` .