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At t=0 second a particle is at point A and moves along ABCDE with uniform speed v=1 π/3 m/s.Both the straight segments AB and DE are along diameter BD of semicircle BCD of radius 1m.Then the instant of time at which the instantaneous velocity of the particle is along the direction of avg velocity from t=0 second to that instant is A.0.5 s B.1 s C.1 π/6 s D.1.5 s?
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At t=0 second a particle is at point A and moves along ABCDE with unif...
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At t=0 second a particle is at point A and moves along ABCDE with unif...
Given Information:
- The particle starts at point A and moves along the path ABCDE.
- The particle moves with a uniform speed of v = π/3 m/s.
- The straight segments AB and DE are along the diameter BD of the semicircle BCD.
- The radius of the semicircle BCD is 1m.

Analysis:
To determine the instant of time at which the instantaneous velocity of the particle is along the direction of the average velocity from t=0 seconds to that instant, we need to find the average velocity vector and the instantaneous velocity vector at different points along the path.

Calculating Average Velocity:
The average velocity can be calculated by dividing the displacement vector by the time interval. Since the particle moves along the path ABCDE, the displacement vector is the vector from point A to point E.

- The displacement vector DE = DEi + DEj (assuming i and j are the unit vectors along the x and y directions, respectively).
- The length of DE can be calculated using Pythagoras' theorem: |DE| = √((1-0)^2 + (1-0)^2) = √2.
- The time interval for the motion from t=0 seconds to that instant is not given, so we'll call it t.

Therefore, the average velocity vector is given by VE_avg = (DE / t) = (√2 / t)i + (√2 / t)j.

Calculating Instantaneous Velocity:
At any point along the path, the instantaneous velocity vector is tangent to the path. Since the straight segments AB and DE are along the diameter BD of the semicircle, the instantaneous velocity vector at points A and E will be along the direction of the average velocity vector.

Conclusion:
Therefore, at the instant of time when the particle reaches point E, the instantaneous velocity vector is along the direction of the average velocity vector from t=0 seconds to that instant. Hence, the correct answer is option D: 1.5 seconds.
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At t=0 second a particle is at point A and moves along ABCDE with uniform speed v=1 π/3 m/s.Both the straight segments AB and DE are along diameter BD of semicircle BCD of radius 1m.Then the instant of time at which the instantaneous velocity of the particle is along the direction of avg velocity from t=0 second to that instant is A.0.5 s B.1 s C.1 π/6 s D.1.5 s?
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At t=0 second a particle is at point A and moves along ABCDE with uniform speed v=1 π/3 m/s.Both the straight segments AB and DE are along diameter BD of semicircle BCD of radius 1m.Then the instant of time at which the instantaneous velocity of the particle is along the direction of avg velocity from t=0 second to that instant is A.0.5 s B.1 s C.1 π/6 s D.1.5 s? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about At t=0 second a particle is at point A and moves along ABCDE with uniform speed v=1 π/3 m/s.Both the straight segments AB and DE are along diameter BD of semicircle BCD of radius 1m.Then the instant of time at which the instantaneous velocity of the particle is along the direction of avg velocity from t=0 second to that instant is A.0.5 s B.1 s C.1 π/6 s D.1.5 s? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At t=0 second a particle is at point A and moves along ABCDE with uniform speed v=1 π/3 m/s.Both the straight segments AB and DE are along diameter BD of semicircle BCD of radius 1m.Then the instant of time at which the instantaneous velocity of the particle is along the direction of avg velocity from t=0 second to that instant is A.0.5 s B.1 s C.1 π/6 s D.1.5 s?.
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