A Particle had a velocity of 18 m/s in the +x direction and 2.4 s later its velocity was m/s in the opposite direction. What was the average acceleration of the particle during this 2.4-s interval?

7 years ago
We assume that the unit vector is along the positive x axis whereas the unit vector is along the positive y axis.
Given:
Initial velocity of the of the particle, .
Final velocity of the particle, .
Time interval over which the velocity vector of particle changes,  .
It is important to note that the negative sign in the final velocity is due to the fact that the velocity vector of the particle is in a direction opposite to the unit vector .
The average acceleration of the particle is defined as:
…… (1)
Where is the change in velocity vector of the particle.
The change in the velocity vector of the particle is given as:

Substitute the given values of the vector in the equation above to obtain as:

Substitute the value of in equation (1) to have

Substitute the given value of in the equation above

The negative sign indicates that the particle is experiencing the acceleration in a direction opposite to the direction of unit vector .
Therefore the magnitude of average acceleration of the particle is 20 m /s2.