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A train moves from rest with acceleratio...

A train moves from rest with acceleration `alpha` and in time `t_(1)` covers a distance `x`. It then decelerates to rest at a constant retardation `beta` for distance `y` in time `t_(2)`. Then -

(A) `(x)/(y)=(beta)/(alpha)`
(B) `(beta)/(alpha)=(t_(1))/(t_(2))`
(C) `x = y`
(D) `(x)/(y)=(betat_(1))/(alphat_(2))`

A

`(x)/(y)=(beta)/(alpha)`

B

`(beta)/(alpha)=(t_(1))/(t_(2))`

C

x =y

D

`(x)/(y)=(betat_(1))/(alphat_(2))`

Text Solution

Verified by Experts

The correct Answer is:
A, B
In the first case,
v =u+at`" ""or", v=0+alphat_(1) = alphat_(1)" ""or",x= (1)/(2)alphat_(1)^(2)`
In the second case, `v^(2)=u^(2)-2as" ""or",0-alpha^(2)t_(1)^(2)-2betay[becauseu=alphat_(1)]`
or, `y = (alpha^(2)t_(1)^(2))/(2beta)`
`:.(x)/(y )=((1)/(2)alphat_(1)^(2))/((alpha^(2)t_(1)^(2))/(2beta))=(beta)/(alpha)`
Also, v =u-at `" ""or", 0=at_(1)-betat_(2)`
or, `(t_(1))/(t_(2))=(beta)/(alpha)`
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