Question

A thief run with a uniform speed of 100m / minute after 1min , a policeman run after the thief to catch him. He goes with a speed of 100m/minute in the first minute and increase his speed by 10m/minute every succeeding minute after how many minute the policeman catch the thief.

Answer 1

In 1 min distance covered by thief = 100 x 1 = 100 m

Therefore gap between thief and police = 100 m

When police starts chasing, his speed in 1st min is same as that of thief.

Therefore ther is no change in the gap.

 

In the 2nd min of chase, speed difference is 10.

Therefore the gap also reduces by 10 m.

Similarly the gap reduces by 20 m in 3rd min.

Now 0 + 10 + 20 + 30 + 40 = 100

Therefore it takes 5 mins for police after he starts to catch the thief.

Therefore policeman takes total of 6 min after the thief starts running.

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Answer 2

Step-by-step explanation:

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here....

theif's speed=100m./min

let time taken by police before catch the thief =x

then the time of theif being catched=x+1

as the police increase his speed=10m./min every minute

his time taken as AP-100,110,120....

Distance of the thief = $time ×speed \\</p><p> =100×(x+1)\\</p><p> =100x+100..........(1)$

Distance covered by police =sum of x terms of AP

=Sn=x/2(2×100+(x-1)10)

=Sn=x(100+5x-5)

=Sn=95x-5x^2...........(2)

as the distance covered by police and theif are same...

=equation 1=equation 2

= 95x-5x^2=100x+100

=5x^2-5x-100=0

= x^2-x-20=0

we solve the equation and get...

x=-4 and x=5

[as the time can't be negative]

= X=5

the time taken by police to catch the thief = 5 minutes