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Race-track turns are often "banked" (tilted inward) so that cars can take them at high speed without skidding. Consider a circular track 2km in length banked at an angle of 20o , and just for fun suppose the track is covered in ice (after a bad storm, lets say). With what speed does a car have to drive in order to make it around the track?

Respuesta :

Answer:33.7 m/s

Explanation:

Given

circumference of track= 2km=2000 m

thus radius r=318.26 m

Banking[tex]=20^{\circ}[/tex]

Let R be the reaction force

Thus [tex]R\sin \theta =\frac{mv^2}{r}[/tex]---1

[tex]R\cos theta =mg[/tex]----2

divide 1 & 2 we get

[tex]\tan \theta =\frac{v^2}{rg}[/tex]

[tex]v=\sqrt{rg\tan \theta }[/tex]

v=33.7 m/s

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