Question

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

Answer 1

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)

___________________________

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)

___________________________

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 2

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.

━━━━━━━━━━━━━━━━━━━━━━━━

☆ 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.

━━━━━━━━━━━━━━━━━━━━━━━━

☆ 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.