The distance travelled in first 8 s, x1= 0 + 1 /2 (5) (8) 2 = 160 m.
At this point the velocity v = u+ at = 0 + (5×8) =40 m s–1
Therefore, the distance covered in last four seconds, x2 = (40 × 4) m =160 m
Thus, the total distance x = x1+x2 = (160+ 160) m = 320 m