Question

a car is slowing to a stop from an initial velocity of 25. 8 m/s if it takes 9.2 to stop what is it’s rate of acceleration

Answer 1

-2.8 m\s^2

Step-by-step explanation:

V1 = 25.8 m/s V2 = 0 m\s T = 9.2 s

a = (V2 - V1) / T = (0-25.8) / 9.2 = -2.8 m\s^2