Question

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?

Answer 1

-26.6 m/s²

Step-by-step 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