Question

A body moving with uniform acceleration describes 12 metre in
third second of its motion and 20 metre in 5th second. Find the
distance covered in the 10th second.​

Answer 1

Sn=u+a(n-1/2)

, For the displacement in the 7th sec, we get,

20=u+a(7-1/2)=u+a(13/2)…………………….. (1)

For the displacement in the 9th sec

24=u+a(9-1/2)=u+a(17/2), ……………………..(2)

(2)-(1) gives 24-20 = 2a or a = 2 m/s^2

Substituting for a in (1), u = 7 m/s

now in

S=ut+1/2at^2

7*15+1/2*2*15*15=105+225=330m