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A train moves from rest with acceleration alpha and in time t_(1) covers a distance x. It then decelerates rest at constant retardation beta for distance y in time t_(2). Then |
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Answer» `(X)/(y) = (BETA)/(alpha)` `alpha=(nu_(0))/(t_(1)),beta=(nu_(0))/(t_(2)):.(beta)/(alpha)=(t_(1))/(t_(2))` Displacement = area under `nu-t` graph `:. X = (1)/(2) t_(1)xxnu_(0)` and `y=(1)/(2)t_(2)xxnu_(0)` Hence `(x)/(y) = (t_(1))/(t_(2))=(beta)/(alpha)` 
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