Given, initial velocity, u=0, acceleration, a=5 ms−2 and time taken, t=8 s.
Let s1 be the distance travelled and v be the final velocity.
From second equation of motion, s=ut+12at2
From first equation of motion, v=u+at⇒v=0+5×8=40 ms−1
Now we have to find the distance covered in time t=12 s
We know, distance=speed×time.
As the speed is constant, after 8 s. i.e, speed =40 ms−1, therefore in the remaining 4 s its covers a distance s2=40×4=160 m
Thus total distnace covered in 12 s, s1+s2=160+160=320 m