question_answer

A highway motorist travels at a constant velocity of\[\text{45 km}{{\text{h}}^{-\text{1}}}\]in a\[\text{3}0\,\text{km}{{\text{h}}^{-\text{1}}}\]zone. A police officer on a motorcycle has been watching from behind a bill board. At the same moment, the speeding motorist passes the bill board; the police officer accelerates uniformly from rest to overtake him. If the acceleration of the police officer is \[10\,\text{km}{{\text{h}}^{-2}}\], how long does he take to reach the motorcyclist?

Answer:

Let the bill board be taken as the origin. Let t be the required time. Let P represent the position where the police officer reaches the motorist. For the motorist. (It is a case of uniform motion). \[v=45km{{h}^{-1}}\] \[\therefore S=45t\,\,km\]                                                         … (1) For the police officer. (It is a case of accelerated motion) \[a=10km{{h}^{-2}}\] \[S=ut+\frac{1}{2}a{{t}^{2}}\] \[S=0+\frac{1}{2}10\times {{t}^{2}}=5{{t}^{2}}km\]                          …. (2) Comparing (2) and (1), we get \[5{{t}^{2}}=45t\,\,or\,\,t=9\,\,hours.\]