Question

A train accelerated uniformly from rest attains a maximum speed of 40 m/s in 20 sec. It travels at this speed for 20 sec and is brought to rest with uniform retardation in 40 sec. The average velocity during this period is:-

Answer 1

case 1:
u=0
v=40m/s
t=20s
v=u+at
40=20a
a=2m/s2
s=ut+1/2at2
s=0+1/2*2*20*20
s=400m

case 2:
v=40m/s
t=20s
s=speed*time=40*20=800m

case 3:
u=40m/s
t=40s
v=0
v=u+at
0=40+40a
-40=40a
a=-1m/s2
s=ut+1/2at2
s=40*40+1/2*(-1)*40*40
s=800m

average velocity=total distance/total time
                               = 400+800+800/20+20+40
                               = 2000/80
                            =25m/s

Hope this helps!
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Answer 2

$\sf\:V = U + at_1 \\ \\ 40 = 0 + a × 20 \\ \\ a = 2 \: m/s^2$

$\sf\implies\: V^2 - U^2 = 2as \\ \\ \sf\implies\: 40^2 - 0 = 2 × 2 × s_1 \\ \\ \large {\boxed {\sf {S_1 = 400 \:m}}}$

$\sf\: S_2 = V × t_2 = 40 × 20 = 800\: m$

V = U + at

0 = 40 + a × 40

a = -1 m/s^2

$\sf\implies\: V^2 - U^2 = 2as \\ \\ \sf\implies\: S_3 = 800 \: m$

Total distance travelled = 400 + 800 + 800 = 2000 m

Total time taken = 80 s

$\sf\implies\:Average\:Velocity = \dfrac {2000}{80} = 25m^{-1}$