Question

if a particle move with velocity v1 for time t2 and v2 for time t2 along a straight line then what is its magnitude of average acceleration?

Answer 1

Answer: The magnitude of the average acceleration is $a = \dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

Explanation:

Given that,

Velocity = $v_{1}$

Time = $t_{1}$

Velocity = $v_{2}$

Time = $t_{1}$

We know that,

Change in velocity in time interval = $\dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

So, the acceleration is

Average acceleration = total change in velocity / total change in time

$a = \dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

Hence, the magnitude of the average acceleration is $a = \dfrac{v_{2}-v_{1}}{t_{2}-t_{1}}$

Answer 2

"Velocity changes from $v_{ 1 }{ \quad to }\quad v_{ 2 }$

Time changes from $t_{ 1 }\quad { to }\quad t_{ 2 }$

We know that,

"Rate of change of velocity" is equal to acceleration.

Acceleration $=\frac { { "Rate\quad of\quad change\quad of\quad velocity" } }{ { Time\quad taken } }$

Time taken $=t_{2}-t_{1}$

"Rate of change of velocity" $=v_{2}-v_{1}$

Hence, Acceleration $\frac{v_2-v_1}{t_2-t_1}$

Both acceleration and velocity is a "vector quantity" as both the quantities indicates direction and magnitude. When distance is integrated we get velocity, when velocity is integrated we get acceleration."