22. (i) a car slows down from 28 m/s to rest in a distance of
88 m. what was its acceleration, assumed constant?

a = -8.912 m/s²

Explanation:

Given,

The initial velocity of the car, u = 28 m/s

The final velocity of the car, v = 0

The distance traveled by car, d = 88 m

The velocity displacement relation is given by the formula

                                          v = d/t

∴                                         t = d/v

Substituting in the above values in the given equation

                                           t = 88/28

                                            = 3.142 s

The acceleration is given by the formula

                                         a = (v-u)/t

                                            = (0 - 28)/3.142

                                            = -8.912 m/s²

The negative sign is that the car is decelerating.

Hence, acceleration a = -8.912 m/s²