• # question_answer A particle moves towards East from a point A to a point B at the rate of 4 km/h and then towards North from B to C at rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively A) $\frac{17}{14}km/h\text{ }and\frac{13}{4}km/h$ B) $\frac{13}{4}km/h\text{ }and\frac{17}{4}km/h$ C) $\frac{17}{9}km/h\text{ }and\frac{13}{9}km/h$ D) $\frac{13}{9}km/h\text{ }and\frac{17}{9}km/h$

Given$AB=12\text{ }km\text{ }({{s}_{1}})$ and $BC=5\,km\,({{s}_{2}})$ Speed from A to$B=4\text{ }km/h$ Time taken, ${{t}_{1}}=\frac{12}{4}=3h$ Speed from B to$C=5\text{ }km/h$ Time taken to complete distance from B to C ${{t}_{2}}=\frac{5}{5}=1h$ $Average\,speed=\frac{Total\,dis\tan ce}{Total\,time}=\frac{{{s}_{1}}+{{s}_{2}}}{{{t}_{1}}+{{t}_{2}}}$ $=\frac{12+5}{3+1}=\frac{17}{4}km/h$ In $\Delta BAC,$ $AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}$ $=\sqrt{144+25}=\sqrt{169},AC=13\,km$ Average velocity $=\frac{\text{Distance}\,\text{AC}}{\text{Total}\,\text{time}}\,=\frac{13}{4}km/h$