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A train accelerating uniormly from rest ...

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?

A

`80//3ms^(-1)`

B

`40 ms^(-1)`

C

`25 ms^(-1)`

D

`30 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`40=(20)a_(1)`,
`a_(1)=2 ms^(-2)`
Further `40=(40)a_(1)`
`therefore a_(2)=1 ms^(-2)`
Therefore acceleration is `2 ms^(-2)` and retardation is `1 ms^(-2)`.
Now, `s_(1)=(1)/(2)a_(1)t_(1)^(2)=(1)/(2)xx2xx(20)^(2)=400 m`
`s_(2)= upsilon_(max) t_(2)=40xx20=800 m`
`s_(3)=(upsilon_(max)^(2))/(2a_(2))=((40)^(2))/(2xx1)=800 m`
Now average velocity `=("Total displacement")/("total time")`
`=(400+800+800)/(20+20+40)=25 ms^(-1)`
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