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Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s?
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Rate of decrease in distance between two objects moving with certain s...
**Given Information:**
- Rate of decrease in distance between two objects moving towards each other = 6 m/s
- Rate of decrease in distance between the objects when moving in some direction = 4 m/s

**Approach:**
Let's assume the speeds of the objects as v1 and v2.

When the objects are moving towards each other, the relative speed is the sum of their individual speeds:
Relative speed = v1 + v2

When the objects are moving in the same direction, the relative speed is the difference of their individual speeds:
Relative speed = v1 - v2

From the given information, we can form the following equations:

1. When the objects are moving towards each other:
Rate of decrease in distance = 6 m/s
Relative speed = v1 + v2

2. When the objects are moving in the same direction:
Rate of decrease in distance = 4 m/s
Relative speed = v1 - v2

**Solution:**

Let's solve the equations to find the values of v1 and v2.

1. When the objects are moving towards each other:
Rate of decrease in distance = 6 m/s
Relative speed = v1 + v2

2. When the objects are moving in the same direction:
Rate of decrease in distance = 4 m/s
Relative speed = v1 - v2

To solve these equations, we can use the method of substitution.

Let's solve equation 1 for v1:
v1 = Rate of decrease in distance - v2

Substitute this value of v1 in equation 2:
Rate of decrease in distance = 4 m/s
Relative speed = (Rate of decrease in distance - v2) - v2
4 = Rate of decrease in distance - 2v2

Now, substitute the given value of rate of decrease in distance:
4 = 6 - 2v2

Simplify the equation:
2v2 = 6 - 4
2v2 = 2
v2 = 1 m/s

Now, substitute this value of v2 in equation 1 to find v1:
v1 = Rate of decrease in distance - v2
v1 = 6 - 1
v1 = 5 m/s

Therefore, the speeds of the objects are:
v1 = 5 m/s
v2 = 1 m/s

Hence, the correct answer is option (A) 5 m/s, 1 m/s.
Community Answer
Rate of decrease in distance between two objects moving with certain s...
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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be.

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.Maximum focal length of eye lens of normal person is

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.Minimum focal length of eye lens of a normal person is

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Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s?
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Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s?.
Solutions for Rate of decrease in distance between two objects moving with certain speed towards each other is 6 m/s and rate of decrease decrease in distance between those objects with same speed when move in some direction is 4 m/s. Then the speeds of objects are . . (A) 5 m/s, 1 m/s (B) 4 m/s, 2 m/s (C) 4 m/s, 1 m/s (D) 5m/s,2m/s? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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