A particle 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