#### To determine

**To find:** When the particle is speeding up and slowing down.

#### Answer

The particle speeds up on the interval (1, 2) and (3, 4).

The particle slows down on the interval (0, 1) and (2, 3).

#### Explanation

The given figure represents the position function of the particle.

It is identified from the graph that the position of the function *s* increases on the intervals (0, 1) and (3, 4) and *s* decreases on the interval (1, 3).

Velocity is positive when the position of the particle increases and the velocity is negative when the position of the particle decreases.

Moreover the acceleration is positive when the graph of the position of the particle is concave upward and it is negative when the curve is convex downward.

Thus, observe the following information.

The velocity is positive on the intervals (0, 1) and (3, 4) and the acceleration is positive on the interval (2, 4).

The velocity is negative on the intervals (1, 3) and the acceleration is negative on the interval (0, 2).

Recall the fact that the particle speeds up when the velocity and acceleration have the same sign and slows down when the velocity and acceleration have the opposite sign.

Therefore, it can be concluded that the particle speeds up on the interval (1, 2) and (3, 4) as *v* and *a* have the same sign and slows down on the interval (0, 1) and (2, 3) as *v* and *a* have the opposite sign.

#### To determine

**To find:** When the particle is speeding up and slowing down.

#### Answer

The particle speeds up on the interval (1, 2) and (3, 4).

The particle slows down on the interval (0, 1) and (2, 3).

#### Explanation

The given figure represents the position function of the particle.

It is identified from the graph that the position of the function *s* is increasing on the interval (3, 4) and s is decreasing on the interval (0, 3).

Velocity is positive when the position of the particle is increasing and the velocity is negative when the position of the particle is decreasing.

Moreover the acceleration is positive when the graph of the position of the particle is concave upward and it is negative when the curve is convex downward.

Thus, observe the following information.

The velocity is positive on the interval (3, 4) and the acceleration is positive on the intervals (0, 1) and (2, 4).

The velocity is negative on the intervals (0, 3) and the acceleration is negative on the interval (1, 2).

Recall the fact that the particle speeds up when the velocity and acceleration have the same sign and slows down when the velocity and acceleration have the opposite sign.

Therefore, it can be concluded that the particle speeds up on the interval (1, 2) and (3, 4) as *v* and *a* have the same sign and slows down on the interval (0, 1) and (2, 3) as *v* and *a* have the opposite sign.