A train accelerating uniformly from rest attains a maximum speed of $40m{s}^{-1}$ in 20 s. It travels at this speed for 20 s and is brought to test with uniform retardation in further 40 s. What is the average velocity during this period?

(i) $v=u+a{t}_1$

$40=0+a\times 20$

$\therefore a=2m{s}^{-2}$

$v^2-u^2=2as$

${40}^2-0=2\times 2\times s_1$

$\therefore s_1=400m$

(ii) $s_2=v\times t_2=40\times 20=800m$

(iii) $v=u+at$

$0=40+a\times 40$

$\therefore a=-1m{s}^{-2}$

$v^2-u^2=2as$

$0^2-{40}^2=2(-1)s_3$

$\therefore s_3=800m$

Total distance travelled 

$=s_1+s_2+s_3$

$=400+800+800=2000m$

Total time taken $=20+20+40=80s$

Average velocity  $=\frac{2000}{80}=25m{s}^{-1}$