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An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).?
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An elevator of height 6 m starts from rest and moves upwards with cons...
**Solution:**
Given,
- Initial velocity, u = 0 m/s (as the elevator starts from rest)
- Acceleration, a = 2 m/s²
- Time taken, t = 20 s
- Height of the elevator, h = 6 m
- Distance travelled by the bolt, S = ?
We need to find the distance travelled by the bolt, S.
**Finding the final velocity of the elevator**
Using the equation of motion, v = u + at, we can find the final velocity of the elevator after 20 seconds.
Here,
- Initial velocity, u = 0 m/s (as the elevator starts from rest)
- Acceleration, a = 2 m/s²
- Time taken, t = 20 s
So, v = u + at
=> v = 0 + 2 × 20
=> v = 40 m/s
Hence, the final velocity of the elevator after 20 seconds is 40 m/s.
**Finding the time taken by the bolt to hit the floor**
The bolt has the same initial velocity as that of the elevator, i.e., 0 m/s. It falls freely under the influence of gravity, with an acceleration of g = 10 m/s².
Let the time taken by the bolt to hit the floor be t1.
Using the equation of motion, h = ut + (1/2)gt², we can find the time taken by the bolt to hit the floor.
Here,
- Initial velocity, u = 0 m/s (as the bolt starts from rest)
- Acceleration, a = g = 10 m/s²
- Height of the elevator, h = 6 m
So, h = ut + (1/2)gt²
=> 6 = 0 × t1 + (1/2) × 10 × t1²
=> 6 = 5t1²
=> t1² = 6/5
=> t1 = √(6/5)
=> t1 = 1.095 s (approx)
Hence, the time taken by the bolt to hit the floor is 1.095 seconds (approx).
**Finding the distance travelled by the bolt**
The distance travelled by the bolt can be found using the formula, S = ut1 + (1/2)gt1², where t1 is the time taken by the bolt to hit the floor.
Here,
- Initial velocity, u = 0 m/s (as the bolt starts from rest)
- Acceleration, a = g = 10 m/s²
- Time taken by the bolt to hit the floor, t1 = 1.095 s (approx)
So, S = ut1 + (1/2)gt1²
=> S = 0 × 1.095 + (1/2) × 10 × (1.095)²
=> S = 6.019 m (approx)
Hence, the distance travelled by the bolt till the time it hits the floor is 6.019 meters (approx).
Given,
- Initial velocity, u = 0 m/s (as the elevator starts from rest)
- Acceleration, a = 2 m/s²
- Time taken, t = 20 s
- Height of the elevator, h = 6 m
- Distance travelled by the bolt, S = ?
We need to find the distance travelled by the bolt, S.
**Finding the final velocity of the elevator**
Using the equation of motion, v = u + at, we can find the final velocity of the elevator after 20 seconds.
Here,
- Initial velocity, u = 0 m/s (as the elevator starts from rest)
- Acceleration, a = 2 m/s²
- Time taken, t = 20 s
So, v = u + at
=> v = 0 + 2 × 20
=> v = 40 m/s
Hence, the final velocity of the elevator after 20 seconds is 40 m/s.
**Finding the time taken by the bolt to hit the floor**
The bolt has the same initial velocity as that of the elevator, i.e., 0 m/s. It falls freely under the influence of gravity, with an acceleration of g = 10 m/s².
Let the time taken by the bolt to hit the floor be t1.
Using the equation of motion, h = ut + (1/2)gt², we can find the time taken by the bolt to hit the floor.
Here,
- Initial velocity, u = 0 m/s (as the bolt starts from rest)
- Acceleration, a = g = 10 m/s²
- Height of the elevator, h = 6 m
So, h = ut + (1/2)gt²
=> 6 = 0 × t1 + (1/2) × 10 × t1²
=> 6 = 5t1²
=> t1² = 6/5
=> t1 = √(6/5)
=> t1 = 1.095 s (approx)
Hence, the time taken by the bolt to hit the floor is 1.095 seconds (approx).
**Finding the distance travelled by the bolt**
The distance travelled by the bolt can be found using the formula, S = ut1 + (1/2)gt1², where t1 is the time taken by the bolt to hit the floor.
Here,
- Initial velocity, u = 0 m/s (as the bolt starts from rest)
- Acceleration, a = g = 10 m/s²
- Time taken by the bolt to hit the floor, t1 = 1.095 s (approx)
So, S = ut1 + (1/2)gt1²
=> S = 0 × 1.095 + (1/2) × 10 × (1.095)²
=> S = 6.019 m (approx)
Hence, the distance travelled by the bolt till the time it hits the floor is 6.019 meters (approx).
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An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).?
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An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).?.
An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).?.
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An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).?, a detailed solution for An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? has been provided alongside types of An elevator of height 6 m starts from rest and moves upwards with constant acceleration of 2 m/s². After 20 s from start, a bolt from the ceiling of elevator detaches and then hits the floor of elevator. If the distance travelled by bolt till the time it hits the floor in metres is S, then find (g = 10 m/s²).? theory, EduRev gives you an
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