A motorcycle racer traveling at 145 km/h loses control in a corner of the track

Question

A motorcycle racer traveling at 145 km/h loses control in a corner of the track and slides across the concrete surface. The combined mass of the rider and bike is 243 kg. The steel of the motorcycle rubs agains the concrete road surface (a) What is the frictional force between the road and the motorcycle and rider? (b) What would be the acceleration of the motorcycle and rider during the wipeout? (c) Assuming there were no barriers to stop the motorcycle and rider, how long would it take the bike and rider to slow to a stop?

Explanation / Answer

a) Coefficient of kinetic friction (k) betweensteel and concrete = 0.30 Mass (m) of rider and bike = 243 kg Now the frictional force between the road and the motorcycleand rider is                   f= mg                    = (0.30)(243kg)(9.8m/s2) = 714.42 N b) The acceleration of the motorcycle and rider during thewipeout is             a = f / m               = (714.42 N) / (243kg)                =-2.94m/s2 Here negative signindicates that the decreasing acceleration of the motorcycle. c) Initial speed (u) of themotorcycle = 145 km/h                                                    = (145)(5/18)m/s                                                    = 40.28 m/s Final speed (v) of themotorcycle = 0 Acceleration (a) of themotorcycle = -2.94m/s2 We know theformula v2 - u2 = 2aS                                   0 - u2 = 2(-2.94m/s2)S                                       S = 275.9 m Time (t) is calculatedas t = u / a                                        = ( 40.28 m/s) / (2.94m/s2)                                        = 13.7 s

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