Question

A thief, after committing a theft runs at a uniform speed of 50m/minute. After 2 minutes, a policeman runs to catch him. He goes 60m in the first minute and increases his speed by 5m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

Answer 1

Hint:Find the distance covered by the thief in 2min. Let t be the time taken to catch the thief. The speed of policemen, which is increasing every min can be taken as an AP. Thus the sum of the AP equals the distance covered by the thief in t minutes.

Complete step-by-step answer:
According to the question it is said that the thief after committing the theft runs at a uniform speed. Now the policeman starts running after him, after 2minutes.
Now the thief covers 50 meter per minute. Hence the distance covered by the thief in 2minutes is,
Distance covered by thief in 2minuted $=50\times 2=100m.$
Hence distance b/w the thief and policeman after 2minutes is 100m.
The policeman starts Charging the thief with an initial speed of 60m/s and increasing his speed by 5m every minute.
Now, Let us consider that the police charged the thief for t minutes.
The thief is having a uniform speed throughout, without any variation. Thus we can say that thief cover a total distance of,
Distance covered by thief = initial distance in 2minutes + (uniform speed $\times $ time)
Distance covered by thief $=\left( 100+50t \right)m.$
The police run in the following pattern 60, 65, 70, 75…….. i.e. the initial speed of a policeman is 60m/min and increases by 5m every minute. Thus we get the following pattern, which is in the form of an AP i.e. arithmetic progression.
Here first term $=a=60$
Common difference $=d=5.$
Number of terms $=t.$
Sum of arithmetic progression is given by the formula,
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Here $n=$ no of terms $=t$ .
$a=60$ and $d=5.$
Thus substitute all three values in the formula of ${{s}_{n}}$
${{S}_{n}}=\dfrac{t}{2}\left[ 2\times 60+\left( t-1 \right)5 \right].$
Now let us simplify it , ${{S}_{n}}=50t+100.$
$\Rightarrow 100+50t=\dfrac{t}{2}\left[ 120+5t-5 \right].$
$\Rightarrow 2\left( 100+50t \right)=t\left[ 5t+115 \right].$
$\Rightarrow 200+100t=5{{t}^{2}}+115t.$
$\Rightarrow 5{{t}^{2}}+115t-100t-200=0.$
$\Rightarrow 5{{t}^{2}}+15t-200=0.$
Divide the entire expression by 5,
$\Rightarrow {{t}^{2}}+3t-40=0.$
Now the above equation is similar to the general quadratic equation $a{{x}^{2}}+bx+c=0$ . Now let us compare both the equation we get,
$a=1,b=3$ and $c=-40.$
Now substitute there values in the quadratic formula and simplify it
\[\begin{align}
  & t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{\text{2a}} \\
 & t=\dfrac{\text{-3}\pm \sqrt{{{3}^{2}}-4\times 1\left( -40 \right)}}{\text{2}\times \text{1}}=\dfrac{\text{-3}\pm \sqrt{9+160}}{\text{2}} \\
 & t=\dfrac{\text{-3}\pm \sqrt{169}}{\text{2}}=\dfrac{\text{-3+13}}{\text{2}} \\
 & t=\dfrac{\text{-3+13}}{2}\text{ or }t=\dfrac{\text{-3-13}}{\text{2}}=\dfrac{-16}{2}=-8 \\
 & \Rightarrow \dfrac{10}{2}=5. \\
\end{align}\]
Time cannot be negative, so $t=-8$ can be neglected. Thus $t=5minutes$ .
Hence the policeman will catch the thief in 5minutes.

Note: You might mistakenly take that the thief increases his speed by 5meter every minute, which is wrong. Read the question carefully to ensure that the data you have taken is correct or not. If the thief increases his speed and the policeman has constant speed, then the policeman can never catch the thief.