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 s for its motion, then find the ratio of average speed to the average velocity of the 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 $17 : 13$
Given,
Dimension of the wall $= 12 m \times 5 m$
Let the total time taken by the insect to reach from one corner to another be $t$
Total distance travelled, $D = A B + B C$
$D = 12 + 5$
$D = 17 m$
$A v e r a g e s p e e d = \frac{T o t a l d i s t a c e}{T o t a l t i m e}$

$A v e r a g e s p e e d = \frac{D}{t}$

$A v e r a g e s p e e d = \frac{17}{t}$

Total displacement, $s = A C$

$s = \sqrt{(A B)^{2} + (B C)^{2}}$

$s = \sqrt{(12)^{2} + (5)^{2}}$

$s = \sqrt{169}$

$s = 13 m$

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

$A v e r a g e v e l o c i t y = \frac{s}{t}$

$A v e r a g e v e l o c i t y = \frac{13}{t}$

$\frac{A v e r a g e s p e e d}{A v e r a g e v e l o c i t y} = \frac{17 / t}{13 / t}$

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

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