Question

Geeta moves along the boundary of a square field of side 20m in 80s. What will be the magnitude of displacement at the end of 200s? Also calculate the average velocity ...
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Answer 1

Ayesha matter kya hai?

Step-by-step explanation:

Now from the question we have,

Side of square field = 20 m

Time to complete one round = 80 s

Perimeter of square = 4 × Side

⇒ Perimeter of square field = 4 × 20

⇒ Perimeter of square field = 80 m

So, she moves 80 m in 80 s

⇒ Speed = 80/80 = 1 m/s

Now at the end of 200s, she travels :

⇒ Total distance = 200 × 1

⇒ Total distance = 200 m

Now number of rounds :

⇒ Distance/Perimeter

⇒ 200/80

⇒ 2.5 rounds.

Now from the attachment we can say that after 2 complete rounds, she will be in the same place i.e. initial point (A) .. But after half more rounds, she will be in point C.

So, displacement = AC

Now using Pythagoras theorem :

✒️ Hypotenuse² = Base² + Altitude²

✒️ AC² = 20² + 20²

✒️ AC² = 400 + 400

✒️ AC² = 800

✒️ AC = √(400 × 2)

✒️ AC = 20√2 m

Therefore, magnitude of displacement = 20√2 m

Now average velocity :

✒️ Av.velocity = Total displacement/Total time

✒️ Av.velocity = 20√2/200

✒️ Av. velocity = 0.14 m/s

Therefore, average velocity = 0.14 m/s