Question

a thief runs with a uniform speed of 100 m/minute. after one minute a policeman runs after the thief to catch him. he goes with a speed of 100 m/minute in the 1st minute and increases his speed by 10 m/minute every succeeding minute.
after how many minutes the policeman will catch the thief?

thank you :)

Answer 1

P------100m-------T-------------------------.C

speed of thief = 100m/ min
initial speed of police(u) = 100m/min
and every 1 minute speed of police increases by 10 m/min .
after one minutes distance covered by thief = 100 × 1 = 100m
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

hence, after 5 minutes police can catch the thief.

Answer 2

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.


This is ur ans hope it will help you
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