Solution:
We have given time = 5 sec
distance = 10 m
so speed = 2m/s
and again
time = 8 sec
distance = 20 m
so speed = 20/8 m/s
so using v = u + at
20/7 = 2 + 8a
a = 6/ 64 m/s²
so int time 10 sec it will cover distance
s = ut +1/2at²
s = 28.125 m
so distance covered in next 2 sec
28.125 m - 20 m = 8 m (approx)
