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Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is?
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Airplanes A and B are flying with constant velocity in the same vertic...
For relative motion perpendicular to line of motion of A
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Airplanes A and B are flying with constant velocity in the same vertic...
Given Information:
- Airplanes A and B are flying in the same vertical plane.
- The angle of airplane A with respect to the horizontal is 30°.
- The angle of airplane B with respect to the horizontal is 60°.
- The speed of airplane A is 100√3 m/s.
- At time t=0s, the observer in airplane A finds airplane B at a distance of 500m.
- The observer in airplane A sees airplane B moving with a constant velocity perpendicular to the line of motion of airplane A.
Approach:
To determine when airplane A just escapes being hit by airplane B, we need to find the time it takes for airplane B to reach a position where it is closest to airplane A's path. This closest distance should be equal to the sum of the widths of the two airplanes, assuming the width of each airplane is negligible compared to the distance between them.
Solution:
Step 1: Calculate the horizontal and vertical components of the velocity of airplane A.
- The speed of airplane A is given as 100√3 m/s.
- The angle of airplane A with respect to the horizontal is 30°.
- The horizontal component of velocity (Vx) can be calculated as: Vx = V * cos(θ)
Vx = 100√3 * cos(30°) = 100√3 * √3/2 = 150 m/s
- The vertical component of velocity (Vy) can be calculated as: Vy = V * sin(θ)
Vy = 100√3 * sin(30°) = 100√3 * 1/2 = 50√3 m/s
Step 2: Calculate the horizontal and vertical components of the velocity of airplane B.
- The angle of airplane B with respect to the horizontal is 60°.
- Since the observer in airplane A sees airplane B moving with a constant velocity perpendicular to the line of motion of airplane A, the horizontal component of velocity (Vx') of airplane B should be equal to Vy.
Vx' = Vy = 50√3 m/s
- The vertical component of velocity (Vy') of airplane B should be equal to -Vx, since it is perpendicular to the line of motion of airplane A.
Vy' = -Vx = -150 m/s
Step 3: Determine the relative velocity of airplane B with respect to airplane A.
- The relative velocity (Vrel) can be calculated as the vector sum of the velocities of airplane A and airplane B.
- The horizontal component of the relative velocity (Vrel_x) can be calculated as: Vrel_x = Vx' - Vx
Vrel_x = 50√3 - 150 = -100√3 m/s
- The vertical component of the relative velocity (Vrel_y) can be calculated as: Vrel_y = Vy' - Vy
Vrel_y = -150 - 50√3 = -150 - 50√3 m/s
Step 4: Calculate the time it takes for airplane B to reach the closest distance to airplane A's path.
- The closest distance is equal to the sum of the widths of the two airplanes,
- Airplanes A and B are flying in the same vertical plane.
- The angle of airplane A with respect to the horizontal is 30°.
- The angle of airplane B with respect to the horizontal is 60°.
- The speed of airplane A is 100√3 m/s.
- At time t=0s, the observer in airplane A finds airplane B at a distance of 500m.
- The observer in airplane A sees airplane B moving with a constant velocity perpendicular to the line of motion of airplane A.
Approach:
To determine when airplane A just escapes being hit by airplane B, we need to find the time it takes for airplane B to reach a position where it is closest to airplane A's path. This closest distance should be equal to the sum of the widths of the two airplanes, assuming the width of each airplane is negligible compared to the distance between them.
Solution:
Step 1: Calculate the horizontal and vertical components of the velocity of airplane A.
- The speed of airplane A is given as 100√3 m/s.
- The angle of airplane A with respect to the horizontal is 30°.
- The horizontal component of velocity (Vx) can be calculated as: Vx = V * cos(θ)
Vx = 100√3 * cos(30°) = 100√3 * √3/2 = 150 m/s
- The vertical component of velocity (Vy) can be calculated as: Vy = V * sin(θ)
Vy = 100√3 * sin(30°) = 100√3 * 1/2 = 50√3 m/s
Step 2: Calculate the horizontal and vertical components of the velocity of airplane B.
- The angle of airplane B with respect to the horizontal is 60°.
- Since the observer in airplane A sees airplane B moving with a constant velocity perpendicular to the line of motion of airplane A, the horizontal component of velocity (Vx') of airplane B should be equal to Vy.
Vx' = Vy = 50√3 m/s
- The vertical component of velocity (Vy') of airplane B should be equal to -Vx, since it is perpendicular to the line of motion of airplane A.
Vy' = -Vx = -150 m/s
Step 3: Determine the relative velocity of airplane B with respect to airplane A.
- The relative velocity (Vrel) can be calculated as the vector sum of the velocities of airplane A and airplane B.
- The horizontal component of the relative velocity (Vrel_x) can be calculated as: Vrel_x = Vx' - Vx
Vrel_x = 50√3 - 150 = -100√3 m/s
- The vertical component of the relative velocity (Vrel_y) can be calculated as: Vrel_y = Vy' - Vy
Vrel_y = -150 - 50√3 = -150 - 50√3 m/s
Step 4: Calculate the time it takes for airplane B to reach the closest distance to airplane A's path.
- The closest distance is equal to the sum of the widths of the two airplanes,
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Airplanes A and B are flying with constant velocity in the same vertic...
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Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is?
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Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is?.
Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is?.
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Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is?, a detailed solution for Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? has been provided alongside types of Airplanes A and B are flying with constant velocity in the same vertical plane at angles 30 and 60 with respect to the horizontal respectively . The speed of A is 100√3 m/s . At time t=0 s; an observer in A finds B at a distance of 500m . This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t=t′ ‚ A just escape being hit by B‚ t’ (in sec) is? theory, EduRev gives you an
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