Question

a thief runs away from a police station with a uniform speed of100 m per min. after a minute a policeman runs behind him to catch. he goes at a speed of 100 m in 1st min and increases his speed by 10 m each succeeding min. after how many min, the policeman catch the thief?​

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 constant speed of 100m per min, distance travelled by him = 100(n+1)

Distance travelled by policeman:

Now the policeman increases his speed by 10m per minute after starting with an initial speed of 100m 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 is an A.P with first term 100 and common difference 10.

Hence, Total distance travelled by police in n minutes = 100 +110 + 120 +......= n/2(2*a+(n-1)d)

Now, Total distance travelled by thief in n minutes = Total distance travelled by policeman in n minutes.

100 (n+1) = n/2 (2*100 + (n-1)10)

100*n + 100 = 100*n + 5n2 - 5*n

5n2 - 5*n - 100 = 0

n2 - n - 20 = 0

n2 + 4*n - 5*n - 20 = 0

n ( n + 4 ) - 5 ( n + 4 ) = 0

( n - 5 )( n + 4 ) = 0

Hence, n = 5 or n = -4.

But n > 0, ( n is the number of minutes )

Therefore n = 5.

Hence, the policeman catches the thief after 5 min

Answer 2

ANSWER

Now, according to the question =>

let the time taken to catch the thief by police be n.

Let the time taken by thief being caught by police be (n+1)

(Since the thief ran 1 minute before the police)

Hence, distance travelled by thief in n+1 seconds is 100(n+1) m

Also we know that =>

Speed of police in 1st second =100m/s

Similaly speed of police in 2nd minute =110m/s

Since the speed got increased by 10m/s we got the AP as >

100,110,120,130,......... An(last term)

So the total distance travelled by police in n minutes will be=>

$\dfrac{n}{2}$ {2a +(n-1)d}

=>$\dfrac{n}{2}${2×100 +(n-1)×10}  - - - - - (i)

Also we know that

Distance travelled by police before catching the thief =100(n+1) - - - (ii)

Now from (i ) and (ii) we get =>

100(n+1) =$\dfrac{n}{2}$ {2×100 +(n-1)×10}

=>100n+100 = 100n +5n(n-1)

On canceling 100n from both side we get=>

100 = 5n(n-1)

=>$5n^2 - 5n - 100$ =0

=>$5(n^2-n-20)$=0

=>$(n^2-n-20)$=0

Since this is an quadratic equation we will have two values for n.

Solving by middle term split methord we get =>

$(n^2-5n+ 4n-20)$ =0

=>n(n-5) +4(n-5) = 0

=>(n-5)(n+4) =0

So,

either

           n = 5

 

              or

            n = -4

(neglected as time can't be negative)

Hence the policeman took 5 minutes to catch the thief.

Remember

1)Sn= $\dfrac{n}{2}$ {2a +(n-1)×d}

-for finding sum till nth terms of an AP.

2)An = a +(n-1)d

-for finding nth term of an AP.