A small body is projected vertically upward with a velocity of 40 m/s. Then the length of path traversed by the body in the time interval from t = 3 s to t = 5 s is 

Time of ascent $=\frac{\mathrm{u}}{\mathrm{g}}=\frac{40}{10} \mathrm{s}=4 \mathrm{s}$

from symmetry of vertical motion, we can write, distance covered in the last second of ascent = distance covered in the first second of descent. 

$\therefore$ length of path traversed $=2\times \frac{1}{2}g.{\text{\hspace{0.17em}t}}^2=10m$