Acceleration during the initial phase:
- Given data:
- Initial velocity (u) = 0 m/s (particle starts from rest)
- Acceleration (a) = 6.0 m/s²
- Time (t) = 8 seconds

- Using the equation of motion v = u + at, where v is the final velocity:
- v = 0 + (6.0)(8) = 48 m/s

Deceleration during the final phase:
- Given data:
- Final velocity (v) = 0 m/s (particle comes to rest)
- Time (t) = 5 seconds

- Using the equation of motion v = u + at:
- 0 = 48 + (a)(5)
- Rearranging the equation, we get:
- -48 = 5a
- Dividing both sides by 5, we get:
- -48/5 = a

Magnitude of the deceleration:
- The magnitude of the deceleration is the absolute value of "a" calculated above.
- Taking the absolute value, we have:
- |-48/5| = 48/5 = 9.6 m/s²

Therefore, the magnitude of the deceleration is 9.6 m/s².

Explanation:
- The particle undergoes two phases: acceleration and deceleration.
- During the acceleration phase, the particle starts from rest and accelerates uniformly at a rate of 6.0 m/s² for 8 seconds.
- Using the equation of motion v = u + at, we find the final velocity at the end of the acceleration phase to be 48 m/s.
- During the deceleration phase, the particle slows down uniformly until it comes to rest.
- We are given the final velocity as 0 m/s and the time taken as 5 seconds.
- Using the equation of motion v = u + at, we can find the deceleration "a".
- Rearranging the equation, we find the magnitude of the deceleration to be 9.6 m/s².
- The negative sign in front of the magnitude indicates that the deceleration is in the opposite direction to the initial motion.
- Therefore, the particle accelerates at 6.0 m/s² for 8 seconds and then decelerates at 9.6 m/s² for 5 seconds until it comes to rest.