Question

An insect moves along the sides of a wall of dimension $12\text{m}\times 5\text{m}$, starting from one corner and reaches the diagonally opposite corner. If the insect takes $10\text{s}$ for its motion, then find the ratio of the average speed to the average velocity of the insect.

• A. $17:12$
• B. $13:5$
• C. $13:17$
• D. $17:13$

Answer 1

The correct option is D $17:13$

Given, the dimension of the wall is $12\text{m}\times 5\text{m}$.
Total time taken, $T=10\text{s}$
Total distance, $d=AB+BC=12+5=17m$


Total displacement, AC = $\sqrt{A{B}^2+B{C}^2}$
(By Pythagoras theorem)
AC = $\sqrt{5^2+{12}^2}$ = $\sqrt{169}$ = $13m$

Average speed
$=\frac{\text{Total distance(d)}}{\text{Total time(T)}}$ = $\frac{17}{10}m{s}^{-1}$
Average velocity $=\frac{Totaldisplacement\left(AC\right)}{Totaltime\left(T\right)}=\frac{13}{10}m{s}^{-1}$
$\therefore \frac{\text{Average speed}}{\text{Average velocity}}=\frac{17}{13}=17:13$