Question

the rate of change of velocity of a body whose initial and final velocity are 2m/s and 4m/s in 1 is​

Answer 1

$\large\maltese\:\underbrace{{\rm{\sf Concept:- }}}$

• The rate of change of velocity is stated as the acceleration.
• The acceleration of the given body can be calculated by using the 1st Kinematical equation, which is outlined as under:-

$\\\leadsto\large\underline{\boxed{\sf v=u+at }}\bigstar$

• where, v, u, a and t represent final velocity, initial velocity, acceleration, and time taken respectively.

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$\large\maltese\:\underbrace{{\rm{\sf Solution:- }}}$

Substituting the values, we get :-

→ 4 = 2 + a × 1

→ 4 = 2 + a

→ a = 2 m/s²

∴ the required answer is 2m/s²

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Kindly note the following points:-

• Negative acceleration which is termed deacceleration/retardation.
• This retardation occurs when the velocity is decreasing.
• Through this body will eventually come to rest and the final velocity will become zero.

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Answer 2

Answer:

• 2 m/s².

Explanation:

We have,

• Initial velocity (u) = 2 m/s
• Final velocity (v) = 4 m/s
• Time taken (t) = 1 sec

To find,

• Acceleration (a)

Solution,

>> Acceleration = Change in velocity/Time taken

Substituting the values:

=> v - u/t

=> 4 - 2/1

=> 2/1

∴ 2 m/s² is the rate of change of velocity or the acceleration of the body.