Question

A train accelerating uniformly from rest attains a maximum speed of 40 ms–1 in 20 s. It travels at this speed for 20 s.and is brought to rest with a uniform retradation in further 40 s.what is the average velocity during this period

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