Two cars are moving on a banked circular path | vedclass

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Question

(B) The maximum speed $v_{max}$ of a car on a banked road with friction is given by the formula: $v_{max} = \sqrt{rg \left( \frac{	an 	heta + \mu}{1

Vedclass · Accepted Answer

$\sqrt{9}: \sqrt{7}$

Vedclass · Answer

(B) The maximum speed $v_{max}$ of a car on a banked road with friction is given by the formula: $v_{max} = \sqrt{rg \left( \frac{	an 	heta + \mu}{1 - \mu 	an 	heta} ight)}$.Given $r = 8 \ m$,$	heta = 45^{\circ}$,so $	an 	heta = 1$.The formula simplifies to $v_{max} = \sqrt{rg \left( \frac{1 + \mu}{1 - \mu} ight)}$.For the first car with $\mu_1 = 0.5$: $v_1 = \sqrt{8g \left( \frac{1 + 0.5}{1 - 0.5} ight)} = \sqrt{8g \left( \frac{1.5}{0.5} ight)} = \sqrt{8g 	imes 3} = \sqrt{24g}$.For the second car with $\mu_2 = 0.4$: $v_2 = \sqrt{8g \left( \frac{1 + 0.4}{1 - 0.4} ight)} = \sqrt{8g \left( \frac{1.4}{0.6} ight)} = \sqrt{8g 	imes \frac{7}{3}} = \sqrt{\frac{56g}{3}}$.The ratio $v_1 : v_2 = \sqrt{24g} : \sqrt{\frac{56g}{3}} = \sqrt{24} : \sqrt{\frac{56}{3}} = \sqrt{72} : \sqrt{56} = \sqrt{9 	imes 8} : \sqrt{7 	imes 8} = 3\sqrt{8} : \sqrt{7	imes 8} = 3 : \sqrt{7} = \sqrt{9} : \sqrt{7}$.

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