Question

A thief runs with the uniform speed of $ 100m{{\min }^{-1}} $. After one minute, a policeman runs after the thief in order to catch him. He goes with a speed of $ 100m{{\min }^{-1}} $ in the first minute and increases his speed by $ 10m{{\min }^{-1}} $. After how many minutes the policeman can catch the thief?

Answer 1

Hint: The distance travelled by a body is the product of velocity of the object and time taken to travel. The velocity of an object is the rate of change of a body’s position with respect to a frame of reference, and it is a function of time. Velocity is similar to a specification of the speed of an object and its direction of motion.

Complete step-by-step answer:
Let the police catches the thief in $ n $ min
As per the question, the thief ran $ 1 $ min before the police
Time taken by the thief just \[n=5\min \] before being caught = $ \left( n+1 \right) $ min
Therefore the Distance travelled by the thief in $ \left( n+1 \right) $ min will be
 $ =100\times \left( n+1 \right)m $
We know the speed of police in $ {{1}^{st}} $ min will be
\[=100m{{\min }^{-1}}\]
The speed of police in $ {{2}^{nd}} $ min will be
\[=110m{{\min }^{-1}}\]
Speed of the police in \[{{3}^{rd}}\] min will be
\[=120m{{\min }^{-1}}\]
And so on.
Therefore \[100,110,120...\]will form an arithmetic progression.
Total distance travelled by the police in $ n $ min =
\[\dfrac{n}{2}\left( 2\times 100+\left( n-1 \right)10 \right)\]
When catching the thief by the police, distance travelled by the thief will be equal to distance travelled by the police
\[100\left( n+1 \right)=\dfrac{n}{2}\left( 2\times 100+\left( n-1 \right)10 \right)\]
Expanding the terms will give,
\[100n+100=100n+\dfrac{n}{2}\left( n-1 \right)10\]
Therefore we can write that,
\[100=n\left( n-1 \right)5\]
After rearranging we will get a quadratic equation,
\[{{n}^{2}}-n-20=0\]
This can be also written as,
\[\left( n-5 \right)\left( n+4 \right)=0\]
Therefore,
\[\begin{align}
  & \left( n-5 \right)=0 \\
 & \left( n+4 \right)=0 \\
\end{align}\]
That is,
\[n=5\]
Or
\[n=-4\]
But negative n will not be possible
So
\[n=5\]
Therefore, the time taken by the policeman in order to catch the thief \[n=5\min \].

Note: An arithmetic progression is a series of numbers in which consecutive terms have a common difference among the terms as a constant value. It is helpful in generalising a set of patterns that we observe in our daily life.