A motorcycle's velocity-time graph is described by four segments. Calculate the acceleration for each segment.

In this interval, the velocity increases from $0 \text{ m/s}$ to $8 \text{ m/s}$.
Initial velocity $u = 0 \text{ m/s}$
Final velocity $v = 8 \text{ m/s}$
Time $t = 5 \text{ s} - 0 \text{ s} = 5 \text{ s}$
The acceleration ($a$) is calculated as:

During this period, the motorcycle undergoes uniform acceleration.

In this interval, the velocity remains constant at $8 \text{ m/s}$.
Since the velocity is constant, the acceleration is $0 \text{ m/s}^2$.
During this period, the motorcycle moves with constant velocity.

In this interval, the velocity decreases from $8 \text{ m/s}$ to $4 \text{ m/s}$.
Initial velocity $u = 8 \text{ m/s}$
Final velocity $v = 4 \text{ m/s}$
Time $t = 25 \text{ s} - 15 \text{ s} = 10 \text{ s}$
The acceleration ($a$) is calculated as:

During this period, the motorcycle undergoes uniform deceleration.

In this interval, the velocity remains constant at $4 \text{ m/s}$.
Since the velocity is constant, the acceleration is $0 \text{ m/s}^2$.
During this period, the motorcycle moves with constant velocity.

In this interval, the velocity decreases from $4 \text{ m/s}$ to $0 \text{ m/s}$.
Initial velocity $u = 4 \text{ m/s}$
Final velocity $v = 0 \text{ m/s}$
Time $t = 40 \text{ s} - 30 \text{ s} = 10 \text{ s}$
The acceleration ($a$) is calculated as:

During this period, the motorcycle undergoes uniform deceleration and comes to a stop.

In this interval, the velocity remains constant at $0 \text{ m/s}$.
Since the velocity is constant and zero, the motorcycle is at rest.

Summary of motion:

• $0 \text{ s}$ to $5 \text{ s}$: Uniform acceleration of $\boxed{\text{1.6 m/s}^2}$.
• $5 \text{ s}$ to $15 \text{ s}$: Constant velocity of $8 \text{ m/s}$ (acceleration $0 \text{ m/s}^2$).
• $15 \text{ s}$ to $25 \text{ s}$: Uniform deceleration of $\boxed{\text{-0.4 m/s}^2}$.
• $25 \text{ s}$ to $30 \text{ s}$: Constant velocity of $4 \text{ m/s}$ (acceleration $0 \text{ m/s}^2$).
• $30 \text{ s}$ to $40 \text{ s}$: Uniform deceleration of $\boxed{\text{-0.4 m/s}^2}$.
• $40 \text{ s}$ to $45 \text{ s}$: At rest (velocity $0 \text{ m/s}$, acceleration $0 \text{ m/s}^2$).