Question

The two ends of a train moving with uniform acceleration pass a certain point with velocities "u" & "v". The velocity with which the middle portion of the train passes the same point is -???????????????????????????????????????????????????????????????????????????????????????????????? Plz help:)

Answer 1

\[\frac {u+v}{2}\]?

Answer 2

Its answer is \[\LARGE \color{red}{\sqrt{\frac{u^2+v^2}{2}.}}\]

Answer 3

hm.

Answer 4

hey @mathslover plz help kar yar:)

Answer 5

wait

Answer 6

Hmmmmmmmmmmmmmm.....................................
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Answer 7

@A.Avinash_Goutham @KingGeorge Plz help:)

Answer 8

@maheshmeghwal9


\[v^2=u^2+2as\]
\[as=\frac{v^2-u^2}{2}\]
let x be the velocity of the middle part of train..

\[x^2=u^2+as\]
\[x=\sqrt{u^2+as}\]
\[x=\sqrt{u^2+\frac{v^2-u^2}{2}}\]
\[x=\sqrt{\frac{v^2+u^2}{2}}\]

Answer 9

got it ?

Answer 10

have to go thanks ..

Answer 11

no I didn't gt it:/

Answer 12

how did u write
x^2=u^2+as

Answer 13

I men why?

Answer 14

mean*

Answer 15

x^2 = u^2 + 2*a*s/2 ... right ?

Answer 16

s/2 = half of length of train ...

Answer 17

understood now ?

Answer 18

Why did u apply this formula?
v^2=u^2+2as.

Answer 19

& I think "s" is distance.
How can it be length of the train?

Answer 20

@mathslover

Answer 21

does this make clear to you ? @maheshmeghwal9

Answer 22

very much clear dude.
thanx a lot:)

Answer 23

my pleasure @maheshmeghwal9

Answer 24

^_^ @mathslover

Answer 25

u can close this question now..

Answer 26

ya I forgot that