The minimum speed at which a car can leave the road at the top is 47.47 m/s
Following are the calculations for the minimal speed at which a vehicle can veer off the road;
The hill has a radius of 230 m
Fc = Fn
[tex]\frac{mv^{2} }{r} = mg \\\frac{v^{2} }{r} =g\\v^{2} =rg \\v=\sqrt{rg}[/tex]
where;
The curving road's radius is 230 meters, and
the acceleration caused by gravity is 9.8 meters per second.
[tex]v = \sqrt{230* 9.8}[/tex]
r = 47.47
As a result, 47.47 m/s is the lowest speed at which a vehicle can exit the road at the top.
Find the approximate centripetal acceleration caused by gravity, g, in meters per second squared, and the radius of the vertical circular motion, r, in meters. Step 2: Using the formula vmin=rg v min = r g, get the object's minimum speed (the speed at the top of the circle).
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