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Answers
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✤ Required Answer:
✒ GiveN:
- Initial velocity = 20 m/s
- Retardation = 4 m/s²
- Distance covered = 42 m
✒ To FinD:
- Final velocity of the body.....?
- Time taken of the body....?
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✤ How to Solve?
This question can be solved using the three equations of motion which is used when body undergoes uniform acceleration or retardation. These equations are:
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
Here, v = final velocity, u = initial velocity, s = distance covered, t = time taken, a = acceleration
♠️ So, By using these equations, let's solve the above question....
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✤ Solution:
We have,
- s = 42 m
- u = 20 m/s
- a = -4 m/s²
By using 3rd equation of motion,
➝ v² = u² + 2as
➝ v² = 20² + 2 × (-4) × 42
➝ v² = 400 + (-336)
➝ v² = 64
➝ v = 8 m
[Neglecting v = -8 m/s]
✒ Therefore, Final velocity of the body = 8m/s
Now, finding the time taken.
Now we also have,
- v = 8 m/s
So, For easier calculation we will use 1st equation of motion
By using 1st equation of motion,
➝ v = u + at
➝ 8 = 20 + (-4)t
➝ 4t = 20 - 8
➝ 4t = 12
➝ t = 3s
✒ Time taken by the body = 3s
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Answer:
Explanation:
Question,
A body moving with a velocity of 20 m/s undergoes a retardation of 4 m/s until it covers a distance of 42 m. Find a. Final velocity b. Time taken.
Solution,
Here, we have
Initial velocity, u = 20 m/s
Retardation, a = - 4 m/s²
Distance covered, d = 42 m.
To Find,
a. Final velocity, v = ?
b. Time taken, t = ?
According to the 3rd equation of motion,
We know that
v² - u² = 2as
So, putting all the values, we get
v² - u² = 2as
⇒ v² - (20)² = 2 × (- 4) × 42
⇒ v² - 400 = - 336
⇒ v² = - 336 + 400
⇒ v² = 64
⇒ v = - 8, 8. (As speed can't be negative)
⇒ v = 8 m/s.
Hence, the final velocity of the body is 8 m/s.
Now, we will find time taken,
According to the 1st equation of motion,
We know that,
v = u + at
So, putting all the values, we get
v = u + at
⇒ 8 = 20 + (- 4) × t
⇒ 8 - 20 = - 4t
⇒ - 12 = - 4t
⇒ 12/4 = t
⇒ t = 3 seconds.
Hence, the time taken by the particle is 3 seconds.