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aatish agrawal Grade: Upto college level
        

A speeder in an automobile passes a stationary policeman who is hiding behind a hill board with a motorcycle.After a 2 sec. delay the policeman accelerates to his maximum speed of 150km/hr in 12sec and catches the speeder 1.5 km away from the hill board. calculate  the speed of speeder in km/hr.

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

maximum velocity of policeman = 150KM/hr = 125/3m/s


 for policeman , applying V = U+at  upto that instant when it has reached its maximum velocity...


   125/3 = 0+a(12)


 a(accleration)= 125/36 m/s2 


 distance covered by policeman upto this instant = 1/2 at2 = 250m = 0.25km


total distance moved by policeman upto the instant when he catches the speeder is 1.5Km (given )


   distance covered with constant velocty = 1.5-0.25 = 1.25KM = 1250m


 V = D/t


 t = 1250/125/3 = 30sec


it means , total time by policeman to catch speeder = 30+12=42sec


now , policeman starts his journey 2 sec later so , total time for speeder = 42+2 = 44sec


now let speeder moves with U , then distance covered in 44sec = 44U = 1.5km


   44u = 1500


      u = 34.09m/s


 

6 years ago
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