An insect moves along the sides of a wall of dimensions 12m $\times$ 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 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 $= A B + B C = 12 + 5 = 17 m$
Total displacement $\sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13 m$ [By pythagoras theorem]

Total displacement = AC
$\frac{Average speed}{Average velocity} = \frac{\frac{Total distance}{time}}{\frac{Total displacement}{time}}$
$= \frac{17}{13} = 17 : 13$

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An insect moves along the sides of a wall of dimensions 12m × 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 average velocity of insect.

The correct option is A 17 : 13 Total distance =AB+BC=12+5=17 m Total displacement √52+122=√169=13 m [By pythagoras theorem] Total displacement = AC Average speedAverage velocity=Total distancetimeTotal displacementtime =1713=17:13

An insect moves along the sides of a wall of dimensions 12m $\times$ 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 average velocity of insect.