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 the rate of 5 km/h. If\[AB=12\]km and\[BC=5\text{ }km,\]there its average speed for its journey from A to C are respectively

A)  \[\frac{17}{9}km/h\,and\,\frac{13}{9}km/h\]

B)  \[\frac{13}{4}km/h\,and\,\frac{17}{4}km/h\]

C)  \[\frac{17}{4}km/h\,and\,\frac{13}{4}km/h\]

D)  \[\frac{13}{9}km/h\,and\,\frac{17}{9}km/h\]

Correct Answer: C

Solution :

                Given, \[{{s}_{1}}=12,{{s}_{2}}=5,{{t}_{1}}=\frac{12}{4}=3h,{{t}_{2}}=\frac{5}{5}=1h\] \[\therefore \]Average speed \[=\frac{{{s}_{1}}+{{s}_{2}}}{{{t}_{1}}+{{t}_{2}}}=\frac{12+5}{3+1}\]                                 \[=\frac{17}{4}km/h\] Average speed \[=\frac{displacement}{total\text{ }time}=\frac{AC}{4}=\frac{13}{4}km/h\]