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A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be?
Verified Answer
A point object starts moving along x-axis with constant velocity 0.5 m...
From the properties of similar triangles, we get,
3/x = 4/1 => x=3/4 m
SO, the object's image will be seen by the observer till it is in AD.
So, length of AD = x + 2 + x= 3/4 + 2 + 3/4 = 2+ 3/2 = 7/2 m
Therefore, time = distance/speed = (7/2) / 0.5 = 7 seconds.
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Most Upvoted Answer
A point object starts moving along x-axis with constant velocity 0.5 m...
Given:
- The point object is moving along the x-axis with a constant velocity of 0.5 meters per second in the positive x-direction from the origin.
- A plane mirror of length 2 meters is placed parallel to the x-axis at a distance of 3 meters from the x-axis and 10 meters from the y-axis.
To Find:
- The time for which an observer positioned at (11, -1) will see the image of the point object in the mirror.
Explanation:
To determine the time for which the observer will see the image of the point object in the mirror, we need to analyze the path of the point object and the position of the observer.
Path of the Point Object:
- The point object is moving along the x-axis with a constant velocity of 0.5 meters per second in the positive x-direction. Therefore, its position at any time t can be given by x = 0.5t.
Position of the Observer:
- The observer is positioned at (11, -1), which means the observer is at a distance of 11 meters along the x-axis and 1 meter below the x-axis.
Distance between the Point Object and the Observer:
- The distance between the point object and the observer can be calculated using the distance formula: d = √((x2 - x1)² + (y2 - y1)²).
- Substituting the values, we have d = √((11 - 0.5t)² + (-1 - 0)²).
Time for which the Observer will See the Image:
- The observer will see the image of the point object when the distance between them is equal to the distance between the mirror and the x-axis (which is 3 meters).
- Setting the distance between the point object and the observer equal to 3 meters, we have √((11 - 0.5t)² + (-1 - 0)²) = 3.
- Squaring both sides of the equation, we get (11 - 0.5t)² + 1 = 9.
- Expanding the equation, we have 121 - 11t + 0.25t² + 1 = 9.
- Simplifying the equation, we get 0.25t² - 11t + 113 = 0.
Using Quadratic Formula to Solve the Equation:
- The equation 0.25t² - 11t + 113 = 0 is a quadratic equation of the form at² + bt + c = 0, where a = 0.25, b = -11, and c = 113.
- Using the quadratic formula t = (-b ± √(b² - 4ac)) / 2a, we can find the values of t.
- Substituting the values, we get t = (-(-11) ± √((-11)² - 4(0.25)(113))) / (2(0.25)).
- Simplifying the equation, we get t = (11 ± √(121 - 113)) / 0.5.
- Further simplifying, we get t = (11 ± √8) / 0.
- The point object is moving along the x-axis with a constant velocity of 0.5 meters per second in the positive x-direction from the origin.
- A plane mirror of length 2 meters is placed parallel to the x-axis at a distance of 3 meters from the x-axis and 10 meters from the y-axis.
To Find:
- The time for which an observer positioned at (11, -1) will see the image of the point object in the mirror.
Explanation:
To determine the time for which the observer will see the image of the point object in the mirror, we need to analyze the path of the point object and the position of the observer.
Path of the Point Object:
- The point object is moving along the x-axis with a constant velocity of 0.5 meters per second in the positive x-direction. Therefore, its position at any time t can be given by x = 0.5t.
Position of the Observer:
- The observer is positioned at (11, -1), which means the observer is at a distance of 11 meters along the x-axis and 1 meter below the x-axis.
Distance between the Point Object and the Observer:
- The distance between the point object and the observer can be calculated using the distance formula: d = √((x2 - x1)² + (y2 - y1)²).
- Substituting the values, we have d = √((11 - 0.5t)² + (-1 - 0)²).
Time for which the Observer will See the Image:
- The observer will see the image of the point object when the distance between them is equal to the distance between the mirror and the x-axis (which is 3 meters).
- Setting the distance between the point object and the observer equal to 3 meters, we have √((11 - 0.5t)² + (-1 - 0)²) = 3.
- Squaring both sides of the equation, we get (11 - 0.5t)² + 1 = 9.
- Expanding the equation, we have 121 - 11t + 0.25t² + 1 = 9.
- Simplifying the equation, we get 0.25t² - 11t + 113 = 0.
Using Quadratic Formula to Solve the Equation:
- The equation 0.25t² - 11t + 113 = 0 is a quadratic equation of the form at² + bt + c = 0, where a = 0.25, b = -11, and c = 113.
- Using the quadratic formula t = (-b ± √(b² - 4ac)) / 2a, we can find the values of t.
- Substituting the values, we get t = (-(-11) ± √((-11)² - 4(0.25)(113))) / (2(0.25)).
- Simplifying the equation, we get t = (11 ± √(121 - 113)) / 0.5.
- Further simplifying, we get t = (11 ± √8) / 0.
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A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be?
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A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be?.
A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A point object starts moving along x-axis with constant velocity 0.5 metre per second in positive x direction from origin a plane mirror of length 2 metre is placed parallel to x axis at a distance 3 from x axis and at a distance of 10 metre from y axis as shown in the figure time for which an observer positioned at 11, minus one will see image of point object in the mirror will be?.
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