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

`19.2m//s` .

Explanation

Here, `2 x = 1.5 m, x = (1.5)/(2) = 0.75 m`
`r = 24.5 m, upsilon =` ?
`h = 0.49m, g = 9.8 m//s^(2)`
`upsilon =sqrt((g r x)/(h)) = sqrt((9.8 xx 24.5 xx 0.75)/(0.49)) = 19.2m//s` .