A cyclist riding a bicycle at a speed of 14 s | vedclass

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(C) For a cyclist taking a turn on a circular road,the angle of inclination $	heta$ with the vertical is given by the formula: $	an 	heta = \frac{v

Vedclass · Accepted Answer

$60$

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(C) For a cyclist taking a turn on a circular road,the angle of inclination $	heta$ with the vertical is given by the formula: $	an 	heta = \frac{v^2}{rg}$.Given values are: speed $v = 14 \sqrt{3} \, m/s$,radius $r = 20 \sqrt{3} \, m$,and acceleration due to gravity $g \approx 10 \, m/s^2$.Substituting these values into the formula:$	an 	heta = \frac{(14 \sqrt{3})^2}{20 \sqrt{3} 	imes 10} = \frac{196 	imes 3}{200 \sqrt{3}} = \frac{588}{200 \sqrt{3}} = \frac{2.94}{\sqrt{3}} \approx \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,we find that $	heta = 60^{\circ}$.

$A$ cyclist riding a bicycle at a speed of $14 \sqrt{3} \, m/s$ takes a turn around a circular road of radius $20 \sqrt{3} \, m$ without skidding. What is his inclination to the vertical (in $^{\circ}$)?

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