A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes,
a policeman runs to catch him. He goes 60 m in first minute and increases his speed by
5 m/minute every suceeding minute. After how many minutes, the policeman will catch
the thief?
[CBSE 2016]
Answers
Answer:
Thief has a lead of 50*2 = 100m
Relative distance covered by cop in 1st min = 60-50 = 10
Relative distance covered by cop in 2nd min = 65-50 = 15
Relative distance covered by cop in 3rd min = 70 -50 = 20
10, 15, 20 ... is an AP whose sum is 100
a= 10 , d = 5
Sum = 100
100 = n/2(2a + (n-1)d) = n/2(20 +(n-1)5 )
100= n/2(20 + 5n -5)
200 = n(15 + 5n)
divide by 5
40 = n(n + 3)
n^2 + 8n - 5n - 40 =0
n(n+8) - 5(n+8)= 0
>> n = 5 as it cant be negative
Therefore cop catches after 5 min + 2min (when cop was at rest) = 7min
Step-by-step explanation:
Answer:
here's your answer mate!
Step-by-step explanation:
According to the question,
According to the question,Distance between thief and police after 2 minutes = 100 m
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes.
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+200
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+2005t
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+2005t 2
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+2005t 2 +15t−200=0
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+2005t 2 +15t−200=0t
According to the question,Distance between thief and police after 2 minutes = 100 mPolice starts chasing thief with initial speed of 60 and increasing his speed by 5 m every minute.Let the police chase the thief for t minutes. Implies, thief covers total 100 + 50t m Police runs with following pattern 60,65,70...This is an Arithmetic Progression with a = 60, d = 5 & number of terms = t t/2(2×60+(t−1)×5)=50t+100t×(120+5t−5)=100t+2005t 2 +15t−200=0t 2
