A thief saw a policeman and started running at a speed of 15 m/s. After 15 seconds, the policeman started running after the thief at a speed of 20 m/s. If the policeman has to catch the thief, the distance (in m) to be covered by the policeman is:

Correct option is C

Let the speed of the thief and the police.

​$$[v_t = 15~\text{m/s}, \qquad v_p = 20~\text{m/s}]$$​​

The thief runs for (15) seconds before the policeman starts.

Distance covered by the thief initially$$d_{\text{start}} = v_t \times t= 15 \times 15= 225~\text{m}$$​​

Relative speed (same direction)

​$$v_{\text{rel}} = v_p - v_t= 20 - 15= 5~\text{m/s}$$​​

Time taken by the policeman to catch the thief

​$$t_{\text{catch}} = \frac{d_{\text{start}}}{v_{\text{rel}}}= \frac{225}{5}= 45~\text{s}$$​​

Distance covered by the policeman

​$$d_p = v_p \times t_{\text{catch}}= 20 \times 45= 900~\text{m}$$​​

​$$\boxed{d_p = 900~\text{m}}$$​​

​$$\boxed{\text{Correct Option: C}}$$​​