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Worked Examples · Example 5

Q.An insect starts from a point A and covers distance AB in 4 seconds. Then covers distance BC in 7 seconds, distance CD in 5 seconds and distance DA in 2 seconds. Calculate the average velocity and average Speed. [The insect moves around quadrilateral ABCD in the order A→B→C→D→A: AB = 5 m (top side), BC = 6 m (right side), CD = 7 m (bottom side), DA = 2 m (left side).]

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★est
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Average speed uses the total path length travelled; average velocity uses the net displacement, which is zero for a full closed loop back to the starting point.

Average speed =Total distance travelledTotal time taken= \dfrac{\text{Total distance travelled}}{\text{Total time taken}}; Average velocity =Net displacementTotal time taken= \dfrac{\text{Net displacement}}{\text{Total time taken}}, where displacement is the straight-line vector from start to end position.

  1. The insect travels around quadrilateral ABCDABCD in the order A→B→C→D→AA\to B\to C\to D\to A, with side lengths AB=5AB=5 m, BC=6BC=6 m, CD=7CD=7 m, DA=2DA=2 m, taking 44 s, 77 s, 55 s, 22 s respectively.
  2. Total distance travelled =AB+BC+CD+DA=5+6+7+2=20= AB+BC+CD+DA = 5+6+7+2 = 20 m.
  3. Total time taken =4+7+5+2=18= 4+7+5+2 = 18 s.
  4. Average speed =Total distanceTotal time=2018=109≈1.11= \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{20}{18} = \dfrac{10}{9} \approx 1.11 m/s. …

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