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  .

unbalanced force

explanation:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer