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The distances covered by a freely falling body in its first, second, third,..., nth seconds of its motion
- a)forms an arithematic progression
- b)forms a geometric progression
- c)do not form any well defined series
- d)form a series corresponding to the difference of square root of the successive natural numbers.
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The distances covered by a freely falling body in its first, second, t...
Distance travelled by a body in nth second is
Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is
Distance travelled by the body in 2nd second is
Distance travelled by the body in 3rd second is
and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.
Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is
Distance travelled by the body in 2nd second is
Distance travelled by the body in 3rd second is
and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.
Most Upvoted Answer
The distances covered by a freely falling body in its first, second, t...
Distance travelled by a body in nth second is
Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is
Distance travelled by the body in 2nd second is
Distance travelled by the body in 3rd second is
and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.
Here, u = 0, a = g
∴ Distance travelled by the body in 1st second is
Distance travelled by the body in 2nd second is
Distance travelled by the body in 3rd second is
and so on.
Hence, the distance covered by a freely falling body in its first, second, third ..... nth second of its motion forms an arithmetic progression.
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Community Answer
The distances covered by a freely falling body in its first, second, t...
Distances covered by a freely falling body in its first, second, third,..., nth seconds of its motion
The motion of a freely falling body is governed by the laws of motion and gravity. When an object falls freely under the influence of gravity, it experiences constant acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
Let's analyze the distances covered by the body in each second of its motion to determine the pattern.
First second:
During the first second of its motion, the body starts from rest and accelerates at a constant rate of 9.8 m/s^2. Using the equation of motion, we can calculate the distance covered in the first second:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 1^2
Distance = 4.9 meters
Second second:
During the second second, the body continues to accelerate at a constant rate of 9.8 m/s^2. However, this time it starts from a velocity gained in the previous second. The distance covered in the second second can be calculated using a similar equation of motion:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 2^2
Distance = 19.6 meters
Third second:
Similarly, during the third second, the body starts with a velocity gained in the previous seconds and accelerates at a constant rate of 9.8 m/s^2. The distance covered in the third second can be calculated as:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 3^2
Distance = 44.1 meters
From the above calculations, we can observe that the distances covered by the body in each second of its motion form an arithmetic progression. The common difference between successive terms is 4.9 meters, which is the distance covered in the first second.
Thus, the correct answer is option 'A' - the distances covered by a freely falling body in its first, second, third,..., nth seconds of its motion form an arithmetic progression.
The motion of a freely falling body is governed by the laws of motion and gravity. When an object falls freely under the influence of gravity, it experiences constant acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
Let's analyze the distances covered by the body in each second of its motion to determine the pattern.
First second:
During the first second of its motion, the body starts from rest and accelerates at a constant rate of 9.8 m/s^2. Using the equation of motion, we can calculate the distance covered in the first second:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 1^2
Distance = 4.9 meters
Second second:
During the second second, the body continues to accelerate at a constant rate of 9.8 m/s^2. However, this time it starts from a velocity gained in the previous second. The distance covered in the second second can be calculated using a similar equation of motion:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 2^2
Distance = 19.6 meters
Third second:
Similarly, during the third second, the body starts with a velocity gained in the previous seconds and accelerates at a constant rate of 9.8 m/s^2. The distance covered in the third second can be calculated as:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 9.8 * 3^2
Distance = 44.1 meters
From the above calculations, we can observe that the distances covered by the body in each second of its motion form an arithmetic progression. The common difference between successive terms is 4.9 meters, which is the distance covered in the first second.
Thus, the correct answer is option 'A' - the distances covered by a freely falling body in its first, second, third,..., nth seconds of its motion form an arithmetic progression.
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The distances covered by a freely falling body in its first, second, third,..., nth seconds of its motiona)forms an arithematic progressionb)forms a geometric progressionc)do not form any well defined seriesd)form a series corresponding to the difference of square root of the successive natural numbers.Correct answer is option 'A'. Can you explain this answer?
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