Question

A body is start from rest has uniform acceleration on 8m/s². The distance travell by it in 5 sec. What is average distance​

Answer 1

Correct Question

• A body is start from rest has uniform acceleration on 8m/s². The distance travell by it in 5 sec. What is the distance covered

Given

• Initial Velocity = 0 m/s
• Acceleration = 8 m/s²
• Time = 5 sec

To Find

• Distance Covered

Solution

● v = u + at [First Equation of Motion]

● v²-u² = 2as [Third Equation of Motion]

✭ Final Velocity :

→ v = u + at

→ v = 0 + 8 × 5

→ v = 0 + 40

→ v = 40 m/s

━━━━━━━━━━━━━━━━━━

✭ Distance Covered :

→ v²-u² = 2as

→ 40²-0² = 2 × 8 × s

→ 1600-0 = 16s

→ 1600/16 = s

→ Distance = 100 m

Answer 2

• Given

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 8 m/s²
• Time (t) = 5 s

• To find

• The distance travelled = ?

• How to solve?

• Firstly, we have to find the final velocity (v) by using the first equation of motion. (v = u + at).

• Then we will find the distance travelled by using the third equation of motion. (v² - u² = 2as)

• Solution -

• v = u + at

⇛ v = 0 + 8 × 5

⇛ v = 40 m/s

• Final velocity = 40 m/s

• v² - u² = 2as

⇒ 40² - 0² = 2 × 8 × s

⇒ 40² = 16 × s

⇒1600/16 = s

⇒ 100 m

• Distance travelled = 100 m