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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
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
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