A thief runs away from a police station with a uniform speed of $100\text{ m/minute}$ . After exactly one minute, a policeman runs after the thief to catch him. The policeman travels at 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 will the policeman catch the thief?

Answer: The policeman will catch the thief in $5$ minutes (or $6$ minutes from the time the thief started running).

Step-by-step solution

Step 1: Express the total distance covered by the thief

Let $n$ be the number of minutes the policeman runs before catching the thief. Since the thief started $1$ minute earlier, the thief runs for a total time of $(n + 1)$ minutes at a constant speed of $100\text{ m/minute}$ .

Step 2: Express the distance covered by the policeman using an AP

The policeman runs $100\text{ m}$ in the 1st minute, $110\text{ m}$ in the 2nd minute, and so on. This forms an Arithmetic Progression with first term $a = 100$ , common difference $d = 10$ , and number of terms $n$ . Applying the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ , we get $\frac{n}{2}[200 + 10n - 10] = \frac{n}{2}(10n + 190) = 5n^2 + 95n$ .

Step 3: Form and simplify the quadratic equation

When the policeman catches the thief, the distances traveled by both must be equal: $5n^2 + 95n = 100n + 100$ . Subtracting $100n + 100$ from both sides gives $5n^2 - 5n - 100 = 0$ . Dividing the entire equation by $5$ yields the simplified quadratic equation $n^2 - n - 20 = 0$ .

Step 4: Factorize and solve for n

We factorize $n^2 - n - 20 = 0$ by splitting the middle term: $n^2 - 5n + 4n - 20 = 0$ , which gives $(n - 5)(n + 4) = 0$ . Since time cannot be negative, $n = -4$ is discarded, leaving $n = 5$ . Thus, the policeman catches the thief in $5$ minutes.

Solve your own maths question free →