A circular road of radius 1000 m has a banki | vedclass

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(A) The formula for the maximum safe speed on a banked road with friction is given by: $V_{max} = \sqrt{Rg \left( \frac{	an 	heta + \mu}{1 - \mu 	a

Vedclass · Accepted Answer

$172$

Vedclass · Answer

(A) The formula for the maximum safe speed on a banked road with friction is given by: $V_{max} = \sqrt{Rg \left( \frac{	an 	heta + \mu}{1 - \mu 	an 	heta} ight)}$ Given: Radius $R = 1000 \ m$,Banking angle $	heta = 45^o$,Coefficient of friction $\mu = 0.5$,Acceleration due to gravity $g = 9.8 \ m/s^2$. Substituting the values: $V_{max} = \sqrt{1000 	imes 9.8 	imes \left( \frac{	an 45^o + 0.5}{1 - 0.5 	an 45^o} ight)}$ Since $	an 45^o = 1$: $V_{max} = \sqrt{9800 	imes \left( \frac{1 + 0.5}{1 - 0.5 	imes 1} ight)}$ $V_{max} = \sqrt{9800 	imes \left( \frac{1.5}{0.5} ight)}$ $V_{max} = \sqrt{9800 	imes 3} = \sqrt{29400} \approx 171.46 \ m/s$. Rounding to the nearest integer,we get $172 \ m/s$.

$A$ circular road of radius $1000 \ m$ has a banking angle of $45^o$. The maximum safe speed of a car having a mass of $2000 \ kg$ will be,if the coefficient of friction between the tyre and the road is $0.5$, . . . . . . $m/s$.

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