a thief runs away from a police station with a uniform speed of100 m per min. after a minute a policeman runs behind him to catch. he goes at a speed of 100 m in 1stmin and increases his speed by 10 m each succeeding min. after how many min, the policeman catch the thief?
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☆☞ Here is ur answer ☜☆
☆☞ Let n be the number of minutes after which policeman catches the thief.
☆☞ then at that instant:
☆☞ Distance travelled by thief = Distance travelled by policeman.
☆☞ Distance travelled by thief:
☆☞ Now the policeman increases his speed by 10m per minute after starting with an initial speed of 100m per minute.
☆☞ Distance travelled by policeman in first minute: 100
☆☞ Distance travelled by policeman in second minute: 110
☆☞ Distance travelled by policeman in third minute: 120
☆☞ Distance travelled by policeman in fourth minute: 130
☆☞ As we can clearly see this is an A.P with first term 100 and common difference 10.
☆☞ Hence, Total distance travelled by police in n minutes = 100 +110 + 120 +......= n/2(2*a+(n-1)d)
☆☞ Now, Total distance travelled by thief in n minutes = Total distance travelled by policeman in n minutes.
◾ 100 (n+1) = n/2 (2*100 + (n-1)10)
◾ 100*n + 100 = 100*n + 5n2 - 5*n
◾ 5n2 - 5*n - 100 = 0
◾ n2 - n - 20 = 0
◾ n2 + 4*n - 5*n - 20 = 0
◾ n ( n + 4 ) - 5 ( n + 4 ) = 0
◾ ( n - 5 )( n + 4 ) = 0
☆☞ Hence, n = 5 or n = -4.
☆☞ But n > 0, ( n is the number of minutes )
☆ Therefore n = 5 ☆
☆☞ Hence, the policeman catches the thief 5 minutes after the policeman starts running.
ENJOY...
GOOD BYE!!!