The distance travelled in first 8 s, x_{1}= 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, x_{2 }= (40 × 4) m =160 m

Thus, the total distance x = x_{1}+x_{2} = (160+ 160) m = 320 m