A body starts from rest with an acceleration of 2m per second ^2 calculate 1 velocity after 5 second ' 2 The distance covered ... In 5 second. .. 3 The distance covered .... By the body while achiving velocity of 40 meter per second
Answers
We have been given that a body starting from rest with an acceleration of 2 m/s²
We have to find i) Velocity after 5 seconds ii) Distance covered in 5 seconds. iii) Distance covered while achieving velocity of 40 m/s
Solution
1) Initial velocity of body = 0 m/s [ As the car starts from rest]
Using 1st equation of motion :
➲ v = u + at
⇒ v = 0 + 2 × 5
⇒ v = 0 + 10
⇒ v = 10 m/s
So, velocity after 5 seconds = 10 m/s (1)
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2) Here u = 0 m/s, a = 2 m/s², t = 5 seconds.
Using 2nd equation of motion :
➲ s = ut + ½ at²
⇒ s = 0 × 5 + ½ × 2 × 5²
⇒ s = 0 + 25
⇒ s = 25 m
So, distance covered by body in 5 s = 25 m (2)
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3) Here, u = 0 m/s, a = 2 m/s² , v = 40 m/s
Using 3rd equation of motion :
➲ v² = u² + 2as
⇒ 40² = 0² + 2 × 2 × s
⇒ 1600 = 4s
⇒ s = 1600/4
⇒ s = 400 m
So, distance covered by body while achieving velocity of 40 m/s = 400 m (3)
Answer:
- Initial Velocity (u) = 0 m/s
- Acceleration (a) = 2m/s²
☆ Velocity after 5 seconds
☢ Using First Equation of Motion :
⇢ v = u + at
⇢ v = 0 + (2 × 5)
⇢ v = 0 + 10
⇢ v = 10 m/s
∴ Velocity after 5 seconds will be 10 m/s.
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☆ The distance covered in 5 seconds
☢ Using Third Equation of Motion :
⇢ v² – u² = 2as
⇢ (10)² – (0)² = 2 × 2 × s
⇢ 100 – 0 = 4s
⇢ 100 = 4s
- Dividing both term by 4
⇢ s = 25 m
∴ Distance covered in 5 seconds is 25 m.
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☆ Distance covered by body while achieving velocity of 40 m/s
☢ Using Third Equation of Motion :
⇢ v² – u² = 2as
⇢ (40)² – (0)² = 2 × 2 × s
⇢ 1600 – 0 = 4s
⇢ 1600 = 4s
- Dividing both term by 4
⇢ s = 400 m
∴ Distance covered by body will be 400 m.