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.

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

Approved