An insect moves along the sides of a wall of dimensions 12 m $\times$ 5 m starting from one corner and reaches the diagonally opposite corner. If the insect takes 2 seconds for its motion, then find the ratio of average speed to the average velocity of insect.

17 : 13

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12 : 5

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13 : 5

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17 : 12

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Solution

The correct option is A
17 : 13

Total distance = AB + BC = 17
Total displacement = AC
$\sqrt{(} A B^{2} + B C^{2} = \sqrt{(} 5^{2} + 12^{2}$ = 13 (By pythagoras theorem)

$\frac{A v e r a g e s p e e d}{A v e r a g e v e l o c i t y} = \frac{\frac{T o t a l d i s t a n c e}{t i m e}}{\frac{T o t a l d i s p l a c e m e n t}{t i m e}}$
$= \frac{T o t a l d i s t a n c e}{T o t a l d i s p l a c e m e n t}$
$= \frac{17}{13} = 17 : 13$ .

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