Question

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

Answer 1

We will first assume the time taken by police to catch the thief and then we will find the distance travelled by thief in $\left( n+1 \right)$ min with the speed of $100\text{ m/minute}$ by using $D=S\times T$, then we will use the condition given in the question that thief increases his speed by $10\text{ m/minute}$every succeeding minute and then form an A.P. and then find the total distance travelled by police and then equate both the distances and find the value of time in which police catches thief.

Complete step-by-step answer :
Let the police catch the thief in $n$ min, now it is given that the thief ran one minute before the police.
Therefore, the time taken by the thief before being caught is $\left( n+1 \right)$ min
It is given that the thief has a speed of $100\text{ m/minute}$,
Now was we know that the $D=S\times T$
Where D = distance, S = speed, T = time.
Now the distance travelled in $\left( n+1 \right)$ min with the speed of $100\text{ m/minute}$ is:
$D=100\times \left( n+1 \right)\text{ m }........\text{Equation 1}\text{.}$
As it is given that the speed of police in the ${{1}^{st}}$ minute = $100\text{ m/minute}$ and then it increases his speed by $10\text{ m/minute}$
So that the speed of police in the ${{2}^{nd}}$ minute = $110\text{ m/minute}$
And the speed of police in the ${{3}^{rd}}$ minute = $120\text{ m/minute}$ and so on
So we see that $100,110,120......$ forms an arithmetic progression and we know that he total sum of terms in the arithmetic progression is: ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ , with $a=100$ where $a$ is the first term and $d=10$ where $d$ is the common difference.
Therefore, the total distance travelled by the police in $n$ min = $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=\dfrac{n}{2}\left[ 2\times 100+\left( n-1 \right)10 \right]\text{ }........\text{ Equation 2}\text{.}$
Now, on catching the thief by police : the distance travelled by thief = distance travelled by the police
Therefore, putting values from equation 1 and equation 2:
$\begin{aligned} & 100\left( n+1 \right)=\dfrac{n}{2}\left( 2\times 100+\left( n-1 \right)10 \right)\Rightarrow 100n+100=\left( \dfrac{n}{2}\times 200 \right)+\left( \dfrac{n}{2}\times \left( n-1 \right)\times 10 \right) \\\ & \Rightarrow 100n+100=100n+\left( \dfrac{n}{2}\times \left( n-1 \right)\times 10 \right) \\\ & \Rightarrow 100=\left( \dfrac{n}{2}\times \left( n-1 \right)\times 10 \right)\Rightarrow 100=n\left( n-1 \right)5 \\\ & \Rightarrow {{n}^{2}}-n-20=0 \\\ \end{aligned}$
Now we will solve this quadratic equation by factorization:

& \Rightarrow {{n}^{2}}-n-20=0 \\\ & \Rightarrow {{n}^{2}}-5n+4n-20=0 \\\ & \Rightarrow n\left( n-5 \right)+4\left( n-5 \right)=0 \\\ & \Rightarrow \left( n-5 \right)\left( n-4 \right)=0 \\\ & \Rightarrow n=5,-4 \\\ \end{aligned}$$ Since the time cannot be negative therefore $n=5$ . **Hence the time taken by the policeman to catch the thief = $5\text{ min}$ .** **Note** : Do not consider the negative value of $n$ as the value of time taken cannot be negative. Take care of the signs while solving the equations as students can make mistakes and also do not forget to mention the units while writing either distance , speed or time.