A particle moves towards east from a point $A$ to a point $B$ at the rate of $4 \mathrm{~km} / \mathrm{h}$…
A particle moves towards east from a point $A$ to a point $B$ at the rate of $4 \mathrm{~km} / \mathrm{h}$ and then towards north from $B$ to $C$ at the rate of $5 \mathrm{~km} / \mathrm{h}$. If $A B=12 \mathrm{~km}$ and $B C=5 \mathrm{~km}$, then its average speed for its journey from $A$ to $C$ and resultant average velocity direct from $A$ to $C$ are respectively
$\frac{17}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{4} \mathrm{~km} / \mathrm{h}$
$\frac{13}{4} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{4} \mathrm{~km} / \mathrm{h}$
$\frac{17}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{13}{9} \mathrm{~km} / \mathrm{h}$
$\frac{13}{9} \mathrm{~km} / \mathrm{h}$ and $\frac{17}{9} \mathrm{~km} / \mathrm{h}$
Solution
Time $T_1$ from $A$ to $B=\frac{12}{4}=3 \mathrm{hrs}$
$T_2$ from $B$ to $C=\frac{5}{5}=1 \mathrm{hrs}$.
Total time $=4 \mathrm{hrs}$.
Average speed $=\frac{17}{4} \mathrm{~km} / \mathrm{hr}$.
Resultant average velocity $=\frac{13}{4} \mathrm{~km} / \mathrm{hr}$.
