At a certain time a particle had a speed of 46 m/s in the positive x direction, and 9.1 s later its speed was 67 m/s in the opposite direction. What was the average acceleration of the particle during this 9.1 s interval

The average acceleration of the particle during this 9.1 s interval is  .

Explanation:

The expression for the average acceleration of the particle is as follows;

Here, is the average acceleration, v is the final speed, u is the initial speed, t is the initial time and t' is the final time.

It is given in the problem that at a certain time a particle had a speed of 46 m/s in the positive x-direction, and 9.1 s later its speed was 67 m/s in the opposite direction.

Put , , t= 0 and t'=9.1 s in the above expression.

Therefore, the average acceleration of the particle during this 9.1 s interval is  .

i think the answer is a, hope this is right, that would be my best guess.

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement