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(D) The maximum safe speed $v$ of a vehicle on a circular road is given by the formula $v = \sqrt{\mu rg}$,where $\mu$ is the coefficient of friction,

Vedclass · Accepted Answer

$5\sqrt{2}\;m/s$

Vedclass · Answer

(D) The maximum safe speed $v$ of a vehicle on a circular road is given by the formula $v = \sqrt{\mu rg}$,where $\mu$ is the coefficient of friction,$r$ is the radius of the curve,and $g$ is the acceleration due to gravity.From this formula,we can see that $v \propto \sqrt{\mu}$.Given initial conditions: $v_1 = 10\;m/s$ and $\mu_1 = \mu$.New conditions: $\mu_2 = \frac{\mu}{2}$.Using the proportionality $v_2 = v_1 \sqrt{\frac{\mu_2}{\mu_1}}$:$v_2 = 10 	imes \sqrt{\frac{\mu/2}{\mu}} = 10 	imes \sqrt{\frac{1}{2}} = \frac{10}{\sqrt{2}}$.Rationalizing the denominator: $v_2 = \frac{10\sqrt{2}}{2} = 5\sqrt{2}\;m/s$.

If the coefficient of friction between the tires and the road is $\mu$,the maximum safe speed is $10\;m/s$. If the coefficient of friction becomes $\mu' = \frac{\mu}{2}$,what will be the new maximum safe speed?

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