Question

A thief runs with a uniform speed of $100m/minute$. After one minute, a policeman runs after the thief to catch him. He goes with a uniform speed of $100m/min$ in the first minute and increases his speed by $10m/minute$ every succeeding minute. After how many minutes the policeman will catch the thief.

Answer 1

Let the police catch the thief in $nmin$

As the thief ran $1min$before the police. So, the time taken by the thief before being caught $=(n+1)min$

Distance travelled by the thief in $(n+1)min=speed\times time==100(n+1)m$

Speed of police in $1stmin=100m/min$

Speed of police in $2ndmin=110m/min$

...$3rdmin=120m/min$.........so on

$100,110,120,$............ this forms an $AP$

Sumof $n$ terms in an $AP=\frac{n}{2}\left[\right(2a+(n-1)d]$

Total distance travelled by the police in $nmin=\frac{n}{2}\left[\right(2\times 100+(n-1)10\left)\right]$

On catching the thief by police,

$distancetraveledbythief=distancetraveledbythepolice$

$100(n+1)=\frac{n}{2}\left[\right(2\times 100+(n-1)10\left)\right]$

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

$100=n(n-1)5$

$n^2-n-20=0$

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

$n-5=0$

$n=5$ OR $n=-4$...(but this is not possible)

so, $n=5$

Hence, the time taken by the policeman to catch the thief $=5min$