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The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is
- a)0.60 ms-2
- b)0.50 ms-2
- c)0.55 ms-2
- d)0.45 ms-2
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its ma...
Maximum acceleration=ω2r
= (0.45)2x3
=0.60ms-2
= (0.45)2x3
=0.60ms-2
Most Upvoted Answer
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its ma...
+ 2) meters.
We can find the velocity and acceleration functions by taking the first and second derivatives of x with respect to time:
velocity v(t) = dx/dt = -3(0.45)sin(0.45t + 2)
acceleration a(t) = d^2x/dt^2 = -3(0.45)^2cos(0.45t + 2)
To find the maximum velocity, we set v(t) equal to zero and solve for t:
0 = -3(0.45)sin(0.45t + 2)
sin(0.45t + 2) = 0
0.45t + 2 = nπ, where n is an integer
t = (nπ - 2)/0.45
To find the maximum acceleration, we set a(t) equal to zero and solve for t:
0 = -3(0.45)^2cos(0.45t + 2)
cos(0.45t + 2) = 0
0.45t + 2 = (n + 0.5)π, where n is an integer
t = [(n + 0.5)π - 2]/0.45
Note that there are infinitely many solutions for both t_max_v and t_max_a, as there are infinitely many values of n.
To find the values of maximum velocity and maximum acceleration, we can substitute the values of t_max_v and t_max_a into the corresponding velocity and acceleration functions:
v_max = -3(0.45)sin(0.45t_max_v + 2)
a_max = -3(0.45)^2cos(0.45t_max_a + 2)
However, since there are infinitely many solutions for t_max_v and t_max_a, we cannot find a unique value for v_max and a_max without additional information.
We can find the velocity and acceleration functions by taking the first and second derivatives of x with respect to time:
velocity v(t) = dx/dt = -3(0.45)sin(0.45t + 2)
acceleration a(t) = d^2x/dt^2 = -3(0.45)^2cos(0.45t + 2)
To find the maximum velocity, we set v(t) equal to zero and solve for t:
0 = -3(0.45)sin(0.45t + 2)
sin(0.45t + 2) = 0
0.45t + 2 = nπ, where n is an integer
t = (nπ - 2)/0.45
To find the maximum acceleration, we set a(t) equal to zero and solve for t:
0 = -3(0.45)^2cos(0.45t + 2)
cos(0.45t + 2) = 0
0.45t + 2 = (n + 0.5)π, where n is an integer
t = [(n + 0.5)π - 2]/0.45
Note that there are infinitely many solutions for both t_max_v and t_max_a, as there are infinitely many values of n.
To find the values of maximum velocity and maximum acceleration, we can substitute the values of t_max_v and t_max_a into the corresponding velocity and acceleration functions:
v_max = -3(0.45)sin(0.45t_max_v + 2)
a_max = -3(0.45)^2cos(0.45t_max_a + 2)
However, since there are infinitely many solutions for t_max_v and t_max_a, we cannot find a unique value for v_max and a_max without additional information.
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The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its ma...
Given equation is compar to standard equation of motion of particle is - x=rcos(wt+π/2) then r=3 and w=0.45 . Now a=rw (Formula) a=3*0.45.=0.60 unit m/s
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The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer?.
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration isa)0.60 ms-2b)0.50 ms-2c)0.55 ms-2d)0.45 ms-2Correct answer is option 'A'. Can you explain this answer?.
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