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A train starts from rest and moves with ...

A train starts from rest and moves with uniform acceleration `alpha` for some time and acquires a velocity `v`. It then moves with constant velocity for some time and then decelerates at rate `beta` and finally comes to rest at the next station. If L is distance between two stations then total time of travel is

A

`(L)/(v) + (v)/(2) ((1)/(alpha) + (1)/(beta))`

B

`(L)/(v) - (v)/(2) ((1)/(alpha) + (1)/(beta))`

C

`(L)/(v) - (v)/(2) ((1)/(alpha) - (1)/(beta))`

D

`(L)/(v) + (v)/(2) ((1)/(alpha) - (1)/(beta))`

Text Solution

Verified by Experts

The correct Answer is:
A

`T= t_(1)+t_(2)+t_(3),` therefore we know `L = (1)/(2)vt_(1)+ Vt_(2)+ (1)/(2)vt_(3)`
`L= (vt_(1))/(2) + Vt_(2)+ (Vt_(3)/(2)`, `t_(2)= (L)/(V) - (10/(2) [ t_(1)+ t_(3)]`….(i)
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