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  • A racing car travels on a track ABCDEFA. ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m.

A racing car travels on a track ABCDEFA. ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is \mu = 0.1. The maximum speed of the car is 50 m/s. Find the minimum time for completing one round.

Answers (1)

To keep the car in circular motion centripetal force is provided by frictional force inward to centre O.

(i) Time is taken by the car to travel from A→B→C

Let s1 be the length of the path

Therefore, s_{1} = \frac{3}{4}[2\pi (2R)]

                       = \frac{3}{4}(4\pi )(100)

                       =300 \pi m

The maximum speed of the car along the circular path is v_{1} = \sqrt{\mu rg}

v_{1} = \sqrt{0.1\times 2\times 100\times 10}=\sqrt{200}=10\sqrt{2}m/s

     = 14.14 m/s

Thus, t_{1} =\frac{ s_{1}}{v_{1}} = \frac{300\pi}{14.14} = 66.62s

(ii) Time taken by the car to travel from C to D & F to A

s2 = CD + FA

    = R + R

    = 200 m

The car will travel with max. speed v2 i.e., 50 m/s

Thus, t_{2} =\frac{ s_{2}}{v_{2}}

               \frac{200}{50}

                 = 4 sec

(iii) Time taken for the path from D to E to F will be

  s_{3}=\frac{1}{4}2\pi R=\frac{1}{4}\times 2 \pi\times 100\times 50 \pi                                   

v_{3}=\sqrt{\mu rg }=\sqrt{0.1\times R\times g}=\sqrt{0.1\times 100 \times 10}=10m/s

               

t_{3} = \frac{s_{3}}{v_{3}}

    \frac{50 \pi}{10}

   = 5\pi sec = 15.7 s

Therefor the total time taken by car =t_{1} +t_{2} + t_{3}

                                                           = 66.62 + 4 +15.7

                                                           = 86.32s

Posted by

infoexpert24

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