Question

I need the solution.....from AP class 10

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