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Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help?
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Two particles A and B are initially at distance d apart. They starts m...
Solution:
Let t be the time taken for the particles to meet.
At time t, the distance travelled by A = vt.
At time t, the distance travelled by B = ut + d (since B is moving along a fixed straight line perpendicular to initial direction of motion of A).
Since the particles meet, the distance travelled by A = the distance travelled by B.
Therefore, vt = ut + d.
Solving for t, we get t = d/(v-u).
Therefore, the particles will meet after time d/(v-u).
Substituting v/(v-u) for t in the expression for distance travelled by A, we get the distance travelled by A before the particles meet.
Distance travelled by A = v * (d/(v-u)) = vd/(v-u).
Substituting u/(v-u) for t in the expression for distance travelled by B, we get the distance travelled by B before the particles meet.
Distance travelled by B = u * (d/(v-u)) + d = ud/(v-u) + d.
Since u^2 < v^2,="" we="" have="" v^2="" -="" u^2="" /> 0.
Therefore, (v-u)(v+u) > 0.
Dividing both sides of the equation vt = ut + d by (v-u), we get d/(v-u) = t - d/(v+u).
Substituting this in the expression for distance travelled by B, we get
Distance travelled by B = ud/(v-u) + d = ut + d = (v+u)t - d/(v+u) = (v+u)d/(v+u) - d/(v+u) = vd/(v+u).
Therefore, the particles will meet after travelling a total distance of vd/(v+u).
Subtracting the distance travelled by A before the particles meet from the total distance travelled by the particles before they meet, we get the distance travelled by B before the particles meet.
Distance travelled by B before the particles meet = vd/(v+u) - vd/(v-u) = 2vd/(v^2-u^2).
Therefore, the particles will meet after travelling a distance of 2vd/(v^2-u^2).
Dividing the distance travelled by B before the particles meet by the time taken for the particles to meet, we get the velocity of B.
Velocity of B = (2vd/(v^2-u^2))/(d/(v-u)) = 2u/(v+u).
Therefore, the particles will meet after time d/(v^2-u^2) and the answer is A.
Let t be the time taken for the particles to meet.
At time t, the distance travelled by A = vt.
At time t, the distance travelled by B = ut + d (since B is moving along a fixed straight line perpendicular to initial direction of motion of A).
Since the particles meet, the distance travelled by A = the distance travelled by B.
Therefore, vt = ut + d.
Solving for t, we get t = d/(v-u).
Therefore, the particles will meet after time d/(v-u).
Substituting v/(v-u) for t in the expression for distance travelled by A, we get the distance travelled by A before the particles meet.
Distance travelled by A = v * (d/(v-u)) = vd/(v-u).
Substituting u/(v-u) for t in the expression for distance travelled by B, we get the distance travelled by B before the particles meet.
Distance travelled by B = u * (d/(v-u)) + d = ud/(v-u) + d.
Since u^2 < v^2,="" we="" have="" v^2="" -="" u^2="" /> 0.
Therefore, (v-u)(v+u) > 0.
Dividing both sides of the equation vt = ut + d by (v-u), we get d/(v-u) = t - d/(v+u).
Substituting this in the expression for distance travelled by B, we get
Distance travelled by B = ud/(v-u) + d = ut + d = (v+u)t - d/(v+u) = (v+u)d/(v+u) - d/(v+u) = vd/(v+u).
Therefore, the particles will meet after travelling a total distance of vd/(v+u).
Subtracting the distance travelled by A before the particles meet from the total distance travelled by the particles before they meet, we get the distance travelled by B before the particles meet.
Distance travelled by B before the particles meet = vd/(v+u) - vd/(v-u) = 2vd/(v^2-u^2).
Therefore, the particles will meet after travelling a distance of 2vd/(v^2-u^2).
Dividing the distance travelled by B before the particles meet by the time taken for the particles to meet, we get the velocity of B.
Velocity of B = (2vd/(v^2-u^2))/(d/(v-u)) = 2u/(v+u).
Therefore, the particles will meet after time d/(v^2-u^2) and the answer is A.
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Two particles A and B are initially at distance d apart. They starts m...
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Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help?
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Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help?.
Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help?.
Solutions for Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? in English & in Hindi are available as part of our courses for NEET.
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ample number of questions to practice Two particles A and B are initially at distance d apart. They starts moving simultaneously with velocity v and u respectively in such a way that A always moves toward B and B moves along a fixed straight line perpendicular to initial direction of motion of A. Then particles will meet after time A. vd/(v^2-u^2) B. ud/(u^2-v^2) C. d/(v^2 u^2)^1/2 D. vd/(v^2 u^2) Ans is given A. But I don't understand how to solve plz help? tests, examples and also practice NEET tests.
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