Question

The velocity of a particle moving along a straight line increased from 3ms-1 to 9ms-1 in 3 seconds. Find the acceleration of the particle.​

Answer 1

Answer:

• 2m/s² is the required answer.

Explanation:

Given:-

• Initial velocity ,u = 3m/s
• Final velocity ,v = 9m/s
• Time taken ,t = 3s

To Find:-

• Acceleration ,a

Solution:-

According to the Question

It is given that velocity of a particle moving along a straight line increased from 3ms-1 to 9ms-1 in 3 seconds . we have to find the acceleration of the particle.

As we know that acceleration is defined as the rate of change in velocity.

• a = v-u/t

where,

• a denote acceleration
• v denote final velocity
• u denote initial velocity
• t denote time taken

Substitute the value we get

→ a = 9-3/3

→ a = 6/3

→ a = 2m/s²

• Hence, the acceleration of the particle is 2m/s².

Answer 2

Answer :-

• 2m/s² is the acceleration of the particle.

Explanation :-

Given :

• Initial velocity (u) = 3 m/s
• Final velocity (v) = 9 m/s
• Time taken (t) = 3 sec

To find :

• Acceleration (a)

Solution :

Acceleration :

• Change in velocity/Time taken

Substituting we get,

⇒ (v - u)/t

⇒ (9 - 3)/3

⇒ 6/3

⇒ 2 m/s²

Therefore,

• Acceleration of the particle is 2 m/s².