Question

A prisoner runs away from police station with a uniform speed of 100m/minute .after one minute a policeman runs behind to catch him.he goes at a speed of 100m/minute and increase his speed 10m/minute each succeding minute .after how many minutes ,the policeman catch the prisoner?

Answer 1

Let n be the number of minutes after which policeman catches the thief
then at that instant:
Distance travelled by thief = Distance travelled by policeman
Distance travelled by thief:
Since the thief moves at a constance speed of 100 m per min, distance travelled by him = 100 (n+1)
Distance travelled by policeman:
Now the policeman incereases his speed by 10 m per minute aftger starting with an initial speed of 100 m per minute.
Distance travelled by policeman in first minute: 100
Distance travelled by policeman in second minute: 110
Distance travelled by policeman in third minute: 120
Distance travelled by policeman in fourth minute: 130
As we can clearly see this an A.P with first term 100 and common difference 10.
Hence, Total distance travelled by police in n minutes = $100+110+120+\dots =\frac{n}{2}(2\times a+(n-1\left)d\right)$
Now, Total distance travelled by thief in n minutes = Total distance travelled by policeman in n minutes.
$100(n+1)=\frac{n}{2}(2\times a+(n-1\left)10\right)100\times n+100=100\times n+5n^2-5\times n5n^2-5\times n-100=0n^2-n-20=0n(n+4)-5(n+4)=0(n-5)(n+4)=0$
Hecem n=5 of n=-4
But n>0, (n is the number of minutes)
Therefore n=5.
Hence, the policeman catches the thief 5 minutes after the policeman starts running.