Question

An object accelerates from rest to a velocity of 22 m/s over 35 m. What was its acceleration? the answer should be 6.91. How do you get that?

Answer 1

Given :-

• An object accelerates from rest to a velocity of 22 m/s over 35 m.

To find :-

• Acceleration of an object

Solution :-

• Initial velocity (u) = 0

• Final velocity (v) = 22m/s

• Distance covered (s) = 35 m

As we know that

→ v² = u² + 2as

Where " v " is final velocity, " u " is Initial velocity, " a " is acceleration and " s " is distance covered by an object.

• This is the third equation of motion.

According to the question

→ v² = u² + 2as

→ (22)² = (0)² + 2 × a × 35

→ 484 = 0 + 70a

→ 484 = 70a

→ a = 484/70

→ a = 6.91 m/s²

Hence,

• Acceleration of an object is 6.91 m/s²

Answer 2

Solution

We have

• Initial velocity (u) = 0
• Final velocity (v) = 22m/s
• Distance covered (s) = 35m

By using third equation of motion

=> v² = u² 2as

★ Putting the values

⟹ (22)² = (0)² × 2a (35)

⟹ 484 = 70a

⟹ a = 484/70

⟹ a = 6.91m²/s

Hence, acceleration of the object is 6.91m²/s.