Explanation

(c) Since, the object accelerates from rest, its initial velocity $u$ is zero. $ie,\ u = 0$
From first equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ 27.5 = 0 + a \times 10$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = 2.75\ {ms}^{- 2}$
Hence, distance covered in first 10 s
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{1} = ut + \frac{1}{2}at^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0 + \frac{1}{2} \times 2.75 \times (10)^{2} = 137.5\ m$
Distance covered in next 10 s with uniform velocity of $27.5\ {ms}^{- 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{2} = 27.5 \times 10 = 275\ m$
Total distance covered
$s = 137.5 + 275 = 412.5\ m$