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The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?
- a)1/v2 + 1/u2 = 4at
- b)1/v2 + 1/u2 = -4at
- c)1/v2 – 1/u2 = 4at
- d)1/v2 – 1/u2 = -4at
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The distance x of a particle moving along a straight line from a fixed...
The velocity of the particle can be found by taking the derivative of the equation for distance with respect to time:
v = d/dt (ax^2 + bx + c)
Using the power rule for differentiation, we can find the derivative of each term separately:
v = 2ax + b
So the velocity of the particle, v, is given by 2ax + b.
The initial velocity, u, can be found by substituting t=0 into the equation for velocity:
u = 2a(0) + b
u = b
Therefore, the initial velocity, u, is equal to b.
v = d/dt (ax^2 + bx + c)
Using the power rule for differentiation, we can find the derivative of each term separately:
v = 2ax + b
So the velocity of the particle, v, is given by 2ax + b.
The initial velocity, u, can be found by substituting t=0 into the equation for velocity:
u = 2a(0) + b
u = b
Therefore, the initial velocity, u, is equal to b.
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The distance x of a particle moving along a straight line from a fixed...
We have, t = ax2 + bx + c ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax2 + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)-1 ……….(2)
Initially, when t = 0 and v = u, let x = x0; hence, from (1) we get,
ax02 + bx0 + c = 0
Or ax02 + bx0 = -c ……….(3)
And from (2) we get, u = 1/(2ax0 + b)
Thus, 1/v2 – 1/u2 = (2ax + b)2 – (2ax0 + b)2
= 4a2x2 + 4abx – 4a2x02 – 4abx0
= 4a2x2 + 4abx – 4a(ax02 – bx0)
= 4a2x2 + 4abx – 4a(-c) [using (3)]
= 4a(ax2 + bx + c)
Or 1/v2 – 1/u2 = 4at
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax2 + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)-1 ……….(2)
Initially, when t = 0 and v = u, let x = x0; hence, from (1) we get,
ax02 + bx0 + c = 0
Or ax02 + bx0 = -c ……….(3)
And from (2) we get, u = 1/(2ax0 + b)
Thus, 1/v2 – 1/u2 = (2ax + b)2 – (2ax0 + b)2
= 4a2x2 + 4abx – 4a2x02 – 4abx0
= 4a2x2 + 4abx – 4a(ax02 – bx0)
= 4a2x2 + 4abx – 4a(-c) [using (3)]
= 4a(ax2 + bx + c)
Or 1/v2 – 1/u2 = 4at
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The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2+ bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?a)1/v2+ 1/u2= 4atb)1/v2+ 1/u2= -4atc)1/v2– 1/u2= 4atd)1/v2– 1/u2= -4atCorrect answer is option 'C'. Can you explain this answer?
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The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2+ bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?a)1/v2+ 1/u2= 4atb)1/v2+ 1/u2= -4atc)1/v2– 1/u2= 4atd)1/v2– 1/u2= -4atCorrect answer is option 'C'. 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 The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2+ bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?a)1/v2+ 1/u2= 4atb)1/v2+ 1/u2= -4atc)1/v2– 1/u2= 4atd)1/v2– 1/u2= -4atCorrect answer is option 'C'. 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 The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2+ bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?a)1/v2+ 1/u2= 4atb)1/v2+ 1/u2= -4atc)1/v2– 1/u2= 4atd)1/v2– 1/u2= -4atCorrect answer is option 'C'. Can you explain this answer?.
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