Question

a car travelling at 22.5m/s decelerate at 2.27m/s². how far does it go in third minutes after the brakes were applied

Answer 1

$$s(t)=v_0t-at^2/2=22.5t-1.135t^2$$

$$d=s(3)-s(2)\\ =(22.5*3-1.135*3^2)-(22.5*2-1.135*2^2)\\ =16.8\:\rm m$$