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A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2 in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip is
- a)1/3 rad.
- b)2/3 rad.
- c)4/3 rad.
- d)2 rad
Correct answer is option 'B'. Can you explain this answer?
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A particle begins to move with a tangential acceleration of constant m...
To solve this problem, we need to understand the relationship between tangential acceleration (at) and total acceleration (a). In a circular path, the total acceleration of a particle is the vector sum of its tangential acceleration and centripetal acceleration.
The centripetal acceleration (ac) is given by the equation:
ac = v^2 / r
where v is the velocity of the particle and r is the radius of the circular path.
Since the tangential acceleration is constant, we can apply Newton's second law to find the relationship between tangential acceleration, total acceleration, and centripetal acceleration:
a^2 = at^2 + ac^2
Given:
at = 0.6 m/s^2
a = 1 m/s^2
Substituting these values into the equation, we get:
1^2 = 0.6^2 + ac^2
1 = 0.36 + ac^2
ac^2 = 1 - 0.36
ac^2 = 0.64
ac = √0.64
ac = 0.8 m/s^2
Now, we can use the centripetal acceleration equation to find the velocity of the particle:
0.8 = v^2 / r
Unfortunately, we don't have enough information to solve for v or r. We would need either the velocity or the radius to find the missing variable.
The centripetal acceleration (ac) is given by the equation:
ac = v^2 / r
where v is the velocity of the particle and r is the radius of the circular path.
Since the tangential acceleration is constant, we can apply Newton's second law to find the relationship between tangential acceleration, total acceleration, and centripetal acceleration:
a^2 = at^2 + ac^2
Given:
at = 0.6 m/s^2
a = 1 m/s^2
Substituting these values into the equation, we get:
1^2 = 0.6^2 + ac^2
1 = 0.36 + ac^2
ac^2 = 1 - 0.36
ac^2 = 0.64
ac = √0.64
ac = 0.8 m/s^2
Now, we can use the centripetal acceleration equation to find the velocity of the particle:
0.8 = v^2 / r
Unfortunately, we don't have enough information to solve for v or r. We would need either the velocity or the radius to find the missing variable.
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A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer?
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A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer?.
A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer?.
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A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A particle begins to move with a tangential acceleration of constant magnitude 0.6ms-2in a circular path. If it slips when its total acceleration becomes 1 ms−2, then the angle through which it would have turned before it started to slip isa)1/3 rad.b)2/3 rad.c)4/3 rad.d)2 radCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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