Question

A truck travelling at 22.5m/ decelerates at 2.27m/s².

How far does it travel during the third second when brakes are applied?

Answer 1

$$d(t)=v_0t+at^2/2$$

$$d=d(3)-d(2)\\ =(v_0*3+a*3^2/2)-(v_0*2+a*2^2/2)\\ =v_0*1+5a*1^2/2\\ =22.5+5*(-2.27/2)=16.8\;\rm m$$