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

Last updated date: 22nd Jul 2024
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Hint: To find the time taken for the policeman to catch the thief, we calculate the distance travelled by the thief and police travelling with uniform speed after ‘n’ minutes. Let us say the thief gets caught after ‘n’ minutes, we equate their distances travelled by then and equate them to solve and find ‘n’.

Given that,
Speed of thief = 100m/min
After 1 min,
Policeman runs with speed = 10 m/min.
Let us consider the policeman who catches the thief in ‘n’ minutes.
The policeman starts running after the thief began to run and finished running 1 min.
Therefore, the time taken by the thief until he gets caught is ‘n+1’ minute.
We know the formula of distance = speed $\times$ time
The speed of thief is given as 100 m/min
Therefore, the thief travels a distance in n+1 minute will be,
Distance of thief = 100 × (n+1) m ------ (1)
According to the question,
The speed of policeman in the first minute is 100 m/min
After 1 minute, the policeman accelerates with 10 m/min,
We know that,
$v = u + at$, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
Here, u = 100 m/min, a = 10 m/min,
So,
The speed of policeman in the second minute will be,
$\Rightarrow 100 + 10 \times 1 = 110$ m/min
The speed of policeman in the third minute will be,
$\Rightarrow 100 + 10 \times 2 = 120$ m/min
And so on, the speed of a policeman in each minute is in the form of an Arithmetic Progression.
We know the formula of sum of n terms in an AP is given as, ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Where a is the first term, n is the number of terms and d is the common difference.
Using this the distance travelled by the policeman in n minutes is,
a = 100, n = n and d = 10
$\Rightarrow \dfrac{n}{2}\left( {2 \times 100 + \left( {n - 1} \right)10} \right)$
The thief has been caught, means that the distance travelled by the police and thief is equal.
$\Rightarrow \dfrac{n}{2}\left( {2 \times 100 + \left( {n - 1} \right)10} \right) = 100\left( {n + 1} \right)$
$\Rightarrow 100n + 5n\left( {n - 1} \right) = 100n + 100$
$\Rightarrow 5n\left( {n - 1} \right) = 100$
$\Rightarrow n\left( {n - 1} \right) = 20$
$\Rightarrow {n^2} - n - 20 = 0$
Now, we will solve this equation by making factors,
$\Rightarrow {n^2} - \left( {5 - 4} \right)n - 20 = 0$
$\Rightarrow {n^2} - 5n + 4n - 20 = 0$
$\Rightarrow n\left( {n - 5} \right) + 4\left( {n - 5} \right) = 0$
$\Rightarrow \left( {n - 5} \right)\left( {n + 4} \right) = 0$
Here, we get two values of n, i.e. n =5 or -4,
But we will take the positive value, as time cannot be negative.

Therefore, the time taken by the policeman to catch the thief is 5 minutes.

Note: In order to solve this type of questions the key is to know the logic behind this problem, i.e. that the distances travelled by the policeman and thief after n minutes should be equal in order for the thief to be caught.
We calculate this by considering they travelled by a variable, n minutes and calculating their distances individually.
It is important to know the formula of distance and the sum of n terms in an AP in order to solve this type of problem.
If the speed per minute is given the value of distance travelled per minute is nothing but the value of speed.