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A particle moves so that its position vector is given by r = cosωtx + sinωty. Where ω is a constant.
Which of the following is true?
- a)Velocity is perpendicular to r and acceleration is directed away from the origin.
- b)Velocity and acceleration both are perpendicular to r
- c)Velocity is acceleration both are parallel to r
- d)Velocity is perpendicular to r and acceleration is directed towards the origin.
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A particle moves so that its position vector is given by r =cosωt...
Explanation:
Given Position Vector:
The position vector of the particle is given by r = costx + sinty, where t is a constant.
Velocity and Acceleration:
- The velocity of the particle can be found by taking the derivative of the position vector with respect to time.
- The velocity vector v = (-sin(t)x + cos(t)y) = -r'
- The acceleration of the particle can be found by taking the derivative of the velocity vector with respect to time.
- The acceleration vector a = (-cos(t)x - sin(t)y) = -v'
Relationship between Velocity and Acceleration:
- The velocity and acceleration vectors are perpendicular to each other if the dot product of the two vectors is zero.
- v.a = (-sin(t)x + cos(t)y) . (-cos(t)x - sin(t)y) = 0
- This implies that the velocity is perpendicular to the acceleration vector.
Direction of Velocity and Acceleration:
- Since the velocity vector v = -r' and acceleration vector a = -v', the velocity is perpendicular to the position vector r.
- Moreover, the acceleration is in the opposite direction of the velocity, which means it is directed towards the origin.
- Therefore, the correct option is D) Velocity is perpendicular to r and acceleration is directed towards the origin.
Given Position Vector:
The position vector of the particle is given by r = costx + sinty, where t is a constant.
Velocity and Acceleration:
- The velocity of the particle can be found by taking the derivative of the position vector with respect to time.
- The velocity vector v = (-sin(t)x + cos(t)y) = -r'
- The acceleration of the particle can be found by taking the derivative of the velocity vector with respect to time.
- The acceleration vector a = (-cos(t)x - sin(t)y) = -v'
Relationship between Velocity and Acceleration:
- The velocity and acceleration vectors are perpendicular to each other if the dot product of the two vectors is zero.
- v.a = (-sin(t)x + cos(t)y) . (-cos(t)x - sin(t)y) = 0
- This implies that the velocity is perpendicular to the acceleration vector.
Direction of Velocity and Acceleration:
- Since the velocity vector v = -r' and acceleration vector a = -v', the velocity is perpendicular to the position vector r.
- Moreover, the acceleration is in the opposite direction of the velocity, which means it is directed towards the origin.
- Therefore, the correct option is D) Velocity is perpendicular to r and acceleration is directed towards the origin.
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Community Answer
A particle moves so that its position vector is given by r =cosωt...
The equation represents UCM hence d
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A particle moves so that its position vector is given by r =cosωtx +sinωty.Where ωis a constant.Which of the following is true?a)Velocity is perpendicular to rand acceleration is directed away from the origin.b)Velocity and acceleration both are perpendicular to rc)Velocity is acceleration both are parallel to rd)Velocity is perpendicular to rand acceleration is directed towards the origin.Correct answer is option 'D'. Can you explain this answer?
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