A train accelerating uniormly from rest attains a maximum speed of `40ms^(-1)` in `20s`. It travels at this speed for `20s` and is brought to rest with uniform retardation i further `40s`. What is the average velocity during this period?

Options

A. `80m//s`
B. `25m//s`
C. `40 m//s`
D. `30m//s`

Explanation

(b) (i) `v=u=at_(1)`
`40=0+axx20`
`a=2m//s^(2)`
`v^(2)-u^(2)=2as`
`40^(@)-0=2xx2s_(1)`
`:. s_(1)=400m`
(ii) `s_(2)=vxxt_(2)=40xx220=800m`
(iii) `v=u+at`
`0=40+axx40`
`:. a=-1m//s^(2)`
`0^(2)=40^(2)=2(-1)s_(3)`
`:. 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 `=2000/80`
`=25m//s`