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A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:?
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A particle moves in a straight line, according to 4a t in()], where ...
To find the velocity of the particle, we need to take the derivative of the position function with respect to time:
dx/dt = 4a(1 + cos(t/a))
This gives us the velocity of the particle at any given time.
To find the acceleration of the particle, we need to take the derivative of the velocity function with respect to time:
d²x/dt² = -4a^2 sin(t/a)
This gives us the acceleration of the particle at any given time. Note that the acceleration is negative when sin(t/a) is negative, which means the particle is slowing down.
We can also find the time at which the particle comes to rest (i.e. when its velocity is zero) by setting dx/dt = 0 and solving for t:
4a(1 + cos(t/a)) = 0
cos(t/a) = -1
t/a = π
t = πa
Therefore, the particle comes to rest at time t = πa.
dx/dt = 4a(1 + cos(t/a))
This gives us the velocity of the particle at any given time.
To find the acceleration of the particle, we need to take the derivative of the velocity function with respect to time:
d²x/dt² = -4a^2 sin(t/a)
This gives us the acceleration of the particle at any given time. Note that the acceleration is negative when sin(t/a) is negative, which means the particle is slowing down.
We can also find the time at which the particle comes to rest (i.e. when its velocity is zero) by setting dx/dt = 0 and solving for t:
4a(1 + cos(t/a)) = 0
cos(t/a) = -1
t/a = π
t = πa
Therefore, the particle comes to rest at time t = πa.
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A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:?
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A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:? 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 in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:?.
A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:? 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 in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a straight line, according to 4a t in()], where x is its the law x=4a(t asin(t/a)) position in meters, t is in sec & a is some constant, then the velocity is zero at:?.
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