A body moving with a uniform acceleration covers 6m and 10m during successive seconds.  Its acceleration is

The average  velocity of the body in the first  1 sec  So distance covered in that time is  6 m,     Therefore  u+a2.1=6  ..........1 If we consider the next second , then the average velocity vavg =(u+a.1+a.1)+u+a2=u+3a2Therefore distance covered in the second second = u+3a2 = 10 ..........2    Solving we get a= 4  m/s2