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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle is
- a)zero
- b)1/√2 m/s2N-W
- c)1/√2 m/s2N-E
- d)1/√2 m/s2S-W
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A particle moves towards east with velocity 5 m/s. After 10 seconds it...
The average acceleration of the particle is not zero since its velocity changed direction.
Using the formula for average acceleration,
average acceleration = (change in velocity) / (time taken)
The change in velocity can be calculated as the final velocity minus the initial velocity.
Initial velocity towards east = +5 m/s
Final velocity towards north = +5 m/s
To calculate the final velocity vector, we can use Pythagoras' theorem.
Magnitude of velocity = 5 m/s
Angle between velocity and north direction = 90 degrees
Using trigonometry,
Magnitude of northward component of velocity = 5 sin(90) = 5
Magnitude of eastward component of velocity = 5 cos(90) = 0
Therefore, the final velocity vector is +5 m/s towards north.
The change in velocity is then:
(change in velocity) = (+5 m/s towards north) - (+5 m/s towards east)
(change in velocity) = +5 m/s towards north
The time taken for the change in velocity to occur is 10 seconds.
Therefore,
average acceleration = (change in velocity) / (time taken)
average acceleration = (+5 m/s towards north) / (10 s)
average acceleration = +0.5 m/s^2 towards north
The answer is (b) 0.5 m/s^2 towards north.
Using the formula for average acceleration,
average acceleration = (change in velocity) / (time taken)
The change in velocity can be calculated as the final velocity minus the initial velocity.
Initial velocity towards east = +5 m/s
Final velocity towards north = +5 m/s
To calculate the final velocity vector, we can use Pythagoras' theorem.
Magnitude of velocity = 5 m/s
Angle between velocity and north direction = 90 degrees
Using trigonometry,
Magnitude of northward component of velocity = 5 sin(90) = 5
Magnitude of eastward component of velocity = 5 cos(90) = 0
Therefore, the final velocity vector is +5 m/s towards north.
The change in velocity is then:
(change in velocity) = (+5 m/s towards north) - (+5 m/s towards east)
(change in velocity) = +5 m/s towards north
The time taken for the change in velocity to occur is 10 seconds.
Therefore,
average acceleration = (change in velocity) / (time taken)
average acceleration = (+5 m/s towards north) / (10 s)
average acceleration = +0.5 m/s^2 towards north
The answer is (b) 0.5 m/s^2 towards north.
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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle isa)zerob)1/√2 m/s2N-Wc)1/√2 m/s2N-Ed)1/√2 m/s2S-WCorrect answer is option 'C'. Can you explain this answer?
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A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle isa)zerob)1/√2 m/s2N-Wc)1/√2 m/s2N-Ed)1/√2 m/s2S-WCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle isa)zerob)1/√2 m/s2N-Wc)1/√2 m/s2N-Ed)1/√2 m/s2S-WCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves towards east with velocity 5 m/s. After 10 seconds its direction changes towards north with same velocity. The average acceleration of the particle isa)zerob)1/√2 m/s2N-Wc)1/√2 m/s2N-Ed)1/√2 m/s2S-WCorrect answer is option 'C'. Can you explain this answer?.
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