Aparticle had a velocity of 19 m/s in the positive x direction and 2.5 s later its velocity was 35 m/s in the opposite direction. what was the average acceleration of the particle during this 2.5 s interval?

-26.6 m/s²

Explanation:

Acceleration: this can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s²

Mathematically, acceleration can be expressed as

a = (v-u)/t Equation 1

a = acceleration, v = velocity, u = initial velocity, t = time.

Given:  u = 19 m/s v = -35 m/s (opposite direction) , t = 2.5 s.

a = (-35-19)/2.5

a = -54/2.5

a = - 26.6 m/s²

Note: a is negative because it is in opposite direction to the initial initial velocity. ( Negative x direction)

Hence the average acceleration = -26.6 m/s²

the body's situation won't change the forces are equal opposit and on the same body