A body moving with uniform acceleration travels $84m$ in $6s$ and $264m$ in $11s$. Find the acceleration of the body.

The correct option is C $4m{s}^{-2}$
$s_1=84m,t_1=6s$
$s_2=264m,t_2=11s$
$s=ut+\frac{1}{2}a{t}^2$
$\therefore s_1=u{t}_1+\frac{1}{2}a{t}_1^2$
$84=6u+(\frac{1}{2}\times a\times 36)$
$84=6u+18a$
$u+3a=14$ -------(i)
$s_2=u{t}_2+\frac{1}{2}a{t}_2^2$
$264=114+(\frac{1}{2}\times a\times 121)$
$u+\frac{11}{2}a=24$ ----- (ii)
Solving (i) and (ii), we get, $a=4m{s}^{-2}$