Question

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

Answer 1

Suppose the policeman catches the thief after t minutes.

Uniform speed of the thief = 50 m/min

∴ Distance covered by thief in (t + 2) minutes = 50 m/min × (t + 2) min = 50 (t + 2) m

The distance covered by the policeman in t minutes is in AP, with 60 and 5 as the first term and the common difference, respectively.

Now,

Distance covered by policeman in t minutes = Sum of t terms

$$\begin{gathered} =\frac{t}{2}\left[2\times 60+\left(t-1\right)\times 5\right] \\ =\frac{t}{2}\left[115+5t\right]\mathrm{m} \end{gathered}$$

When the policeman catches the thief, we have

$$\begin{gathered} \frac{t}{2}\left[115+5t\right]=50\left(t+2\right) \\ 115t+5t^2=100t+200 \\ \Rightarrow 5t^2+15t-200=0 \\ \Rightarrow t^2+3t-40=0 \\ \Rightarrow \left(t+8\right)\left(t-5\right)=0 \end{gathered}$$

So, t = −8 or t = 5

∴ t = 5 (As t cannot be negative)

Thus, the policeman catches the thief after 5 min.