$A$ horizontal curve on a racing track is banked at a $45^o$ angle. When a vehicle goes around this curve at the curve's safe speed (no friction needed to stay on the track),what is its centripetal acceleration?

$A$ road is $10 \text{ m}$ wide. Its radius of curvature is $50 \text{ m}$. The outer edge is above the inner edge by a distance of $1.5 \text{ m}$. This road is most suited for the velocity $[g = 9.8 \text{ m/s}^2]$: (in $\text{ m/s}$)

$A$ car of mass $1000\,kg$ negotiates a banked curve of radius $90\,m$ on a frictionless road. If the banking angle is $45^\circ$,the maximum speed of the car is ............ $m/s$ $[g = 10\,m/s^2]$.

Roads are banked on curves so that

The radius of a curved road on a national highway is $R$. The width of the road is $b$. The outer edge of the road is raised by $h$ with respect to the inner edge so that a car with velocity $v$ can pass safely over it. The value of $h$ is

$A$ railway line is taken round a circular arc of radius $1000 \ m$,and is banked by raising the outer rail $h \ m$ above the inner rail. If the lateral force on the inner rail when a train travels round the curve at $10 \ ms^{-1}$ is equal to the lateral force on the outer rail when the train's speed is $20 \ ms^{-1}$,then the value of $4g \tan \theta$ is equal to: (The distance between the rails is $1.5 \ m$)