# physics

two ends of a train moving with constant acceleration pass a certain point with velocities u and v. what is the velocity with which the middle point of the train passes the same point?
I have asked this question before
you have answered u+v/2. but my mam says it is (u^2+v^2/2)^1/2
please tell me how it comes.

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1. I agree with mam. What you are figuring is the rms velocity.

Here is why:

velocity u is front, then velocity rear, u is
u^2=v^2+2aL where L is the length of the train. so
l= (u^2-v^2)/2a

now at midpoint,
m^2=v^2+2a L/2
m^2=v^2+a (u^2-v^2)/2a
=(u^2+v^2)/2

Sorry for the earlier post.

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2. we know that v2-u2=2as
NOTE :[2 behind u or v is power of them]
3u= tail of train [v=3u]
so (3u)2-u2=2as
9u2-u2=2as
8u2=2as
a=4u2/s
Now at middle of train s'=s/2
v'2-u2=2as'
v'2=u2+2as/2
put the value of a
v'2= u2+2*4u2/s*s/2
v'2=5u2
v'= root5u

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3. siddharth sharma why are you taking taking tail of train as 3u

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posted by vishal
4. Let velocity at mod point is V .

<---------L------->.<---------L ---------->
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
v. V. u

(V^2-v^2)/2a=(u^2-V^2)/2a

So at midpoint velocity = √{(u^2+v^2)/2}

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posted by GOHAN
5. Let velocity at mod point is V .

<---------L------->.<---------L ---------->
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ {train}
v. V. u

(V^2-v^2)/2a=(u^2-V^2)/2a

So at midpoint velocity V = √{(u^2+v^2)/2}

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posted by GOHAN

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