Question

A train accelerated uniformly from rest attains a maximum speed of 40m/s in 20 sec. it travels at this speed for 20 sec and is brought to rest with uniform retardation in 40 secs . the average velocity during this period is ... corrects ans=25 m/s . Anyone can help me since my answer is not coming correct

Answer 1

case 1:
u=0
v=40m/s
t=20s
v=u+at
40=20a
a=2m/s^2
s=ut+1/2at^2
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/s^2
s=ut+1/2at^2
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