Question

At a certain time a particle had a speed of 48 m/s in the positive x direction, and 4.5 s later its speed was 92 m/s in the opposite direction. What was the average acceleration of the particle during this 4.5 s interval?

Answer 1

Answer:

$-31.1 m/s^2$

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is calculated as:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it is important to also take into account the direction of the velocity.

For the particle in this problem, we have:

u = +48 m/s is the initial velocity (positive direction)

v = -92 m/s is the final velocity (negative direction)

t = 4.5 s is the time interval

Therefore, the average acceleration is

$a=\frac{v-u}{t}=\frac{-92-(+48)}{4.5}=-31.1 m/s^2$