A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner.The coefficients of friction between the road and the tires are 0.70 (static)and 0.40 (kinetic).What is the minimum radius of curvature for the corner in order for the car not to skid? A)3.5 $\times$ 10² m B)64 m C)2.1 $\times$ 10² m D)1.6 $\times$ 10² m E)91 m

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Quizplus · Accepted Answer

The minimum radius of curvature can be calculated using the formula $ r = \frac{v^2}{\mu_s g} $, where $ v = 25 \, \text{m/s} $, $ \mu_s = 0.70 $, and $ g = 9.8 \, \text{m/s}^2 $. Plugging in these values gives $ r = \frac{25^2}{0.70 \times 9.8} \approx 91 \, \text{m} $.

A Proud New Jaguar Owner Drives Her Car at a Speed