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A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?
- a)1/v2 = 1/2u2 + 2kt
- b)1/v2 = 1/2u2 – 2kt
- c)1/v2 = 1/u2 – 2kt
- d)1/v2 = 1/u2 + 2kt
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A particle moves in a horizontal straight line under retardation kv3, ...
The correct option is (a) 1/v2 = 1/2u2 + kt.
We can use the equation of motion for uniform retardation, which is given by:
v2 = u2 - 2ax
where v is the velocity at time t, u is the initial velocity, a is the retardation, and x is the displacement at time t.
In this case, the retardation is kv3, so we can substitute a = kv3 into the equation of motion:
v2 = u2 - 2kxv3
We can rearrange this equation to get:
1/v2 = 1/u2 - 2kx/u2v3
We know that x is the displacement at time t, so we can substitute x = vt into the equation above:
1/v2 = 1/u2 - 2kt
Rearranging this equation gives us:
1/v2 = 1/2u2 + kt
Therefore, option (a) is correct.
We can use the equation of motion for uniform retardation, which is given by:
v2 = u2 - 2ax
where v is the velocity at time t, u is the initial velocity, a is the retardation, and x is the displacement at time t.
In this case, the retardation is kv3, so we can substitute a = kv3 into the equation of motion:
v2 = u2 - 2kxv3
We can rearrange this equation to get:
1/v2 = 1/u2 - 2kx/u2v3
We know that x is the displacement at time t, so we can substitute x = vt into the equation above:
1/v2 = 1/u2 - 2kt
Rearranging this equation gives us:
1/v2 = 1/2u2 + kt
Therefore, option (a) is correct.
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Community Answer
A particle moves in a horizontal straight line under retardation kv3, ...
Since the particle is moving in a straight line under a retardation kv3, hence, we have,
dv/dt = -kv3 ……….(1)
Or dv/v3 = -k dt
Or ∫v-3 dv = -k∫dt
Or v-3+1/(-3 + 1) = -kt – c [c = constant of integration]
Or 1/2v2 = kt + c ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u2 = c
Thus, putting c = 1/2u2 in (2) we get,
1/2v2 = kt + 1/2u2
Or 1/v2 = 1/u2 + 2kt
dv/dt = -kv3 ……….(1)
Or dv/v3 = -k dt
Or ∫v-3 dv = -k∫dt
Or v-3+1/(-3 + 1) = -kt – c [c = constant of integration]
Or 1/2v2 = kt + c ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u2 = c
Thus, putting c = 1/2u2 in (2) we get,
1/2v2 = kt + 1/2u2
Or 1/v2 = 1/u2 + 2kt
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A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer?
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A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer?.
A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?a)1/v2 = 1/2u2 + 2ktb)1/v2 = 1/2u2 – 2ktc)1/v2 = 1/u2 – 2ktd)1/v2 = 1/u2 + 2ktCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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