Question

a thief runs with a uniform speed of 100m/min after minutes a police man runs after the thief to catch him he goes with speed of 100m/min in the first minute and increase his speed by10m/min every successding minutes after how many minutes the police man catch the thief . please give me answer for this right answer please

Answer 1

Given
Speed of thief =100m/min

After 1 minutes police start going behind him
In that min the the thief already covered 100m
So policeman have to cover that distance 1st

So

In 1 min thief runs=100m
In that min police man=100m

In next thief=100
Police=110
(so extra distance left=100-10=90)

In 3rd min thief =100m

Policeman =120m

Extra distance left=90-20=70

In 4 min thief=100m

Police man =130

Extra distance =70-30=40

In 5 min thief=100m

Policeman=140m

Extra distance =40-40=0


So in 5 min policeman will catch the thief.


Now using Ap=>

Let police catch the thief after t minutes .
distance covered by thief after t minutes = 

initial position + speed of thief × time =100 + 100t m

distance covered by police after t sec = sum of t terms in AP 
{ where u = 100m/min is the first term and d = 10m/min is the common difference}

= t/2[ 2 × 100 + (t - 1) × 10]

= 100t + 5t² - 5t = 95t + 5t² 


now, when police catches the thief , 

distance covered by thief = distance covered by police 
100 + 100t = 95t + 5t² 
100 + 5t = 5t² => 5t² - 5t - 100 = 0
t² - t -20 = 0 => (t - 5)(t + 4) = 0
t = 5,-4 but t≠ negative so, t = 5 


This is ur ans hope it will help you in case of any doubt comment below 

Answer 2

Let the police catch the thief in ‘n’ minutes
Since the thief ran 1 min before the police start running.
Time taken by the thief before he was caught = (n + 1) min
Distance travelled by the thief in (n+1) min = 100(n+1)m
Given speed of policeman increased by 10m per minute.
Speed of police in the 1st min = 100m/min
Speed of police in the 2nd min = 110m/min
Speed of police in the 3rd min = 120m/min
Hence 100, 110, 120… are in AP
Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]
= (n/2)[2 x 100 +(n – 1)10]
After the thief was caught by the police,
Distance traveled by the thief = distance travelled by the police
100(n+1)= (n/2)[2 x 100 +(n-1)10]
200(n + 1) = n[200 + 10n – 10]
200n + 200= 200n + n(10n – 10 )
200 = n(n – 1)10
n(n – 1) – 20 = 0
n2 – n– 20 = 0
n2 – 5n + 4n – 20 = 0
n(n – 5) + 4(n – 5) = 0
(n – 5) (n+4) = 0
(n – 5) = 0 or (n + 4) = 0
n= 5 or n= -4
Hence n= 5 since n cannot be negative
Therefore the time taken by the policeman to catch the thief = 5minutes

Hope it helps you.....