A particle has an initial velocity of 5.5 ms^-1 due east and a constant acceleration of 1 $m{s}^{-2}$ due west. The distance covered by the particle in sixth second of its motion is

At t= 5 s

$\mathrm{S}=5.5\times 5-\frac{1}{2}\times 1\times {\left(5\right)}^2=15\mathrm{m}$

At t = 6 s

$S=5.5\times 6-\frac{1}{2}\times 1\times {\left(6\right)}^2=15m$

v = u + at  or 0 = 5.5 - t  or  t = 5.5 s

That is the body comes to rest after 5.5 s

At t = 5.5 s

$\mathrm{S}=5.5\times 5.5-\frac{1}{2}\times 1\times {(5.5)}^2=15.125\mathrm{m}$

Distance covered in 6th second

$2\times 0.125=0.25\mathrm{m}$