question_answer

A train moves from rest with acceleration\[\alpha \]and in time \[{{t}_{1}}\]covers a distance x. It then decelerates to rest at constant retardation\[\beta \]for distance y in time \[{{t}_{2}}\] Then

A) \[\frac{x}{y}=\frac{\beta }{\alpha }\]     

B) \[\frac{\beta }{\alpha }=\frac{{{t}_{2}}y}{{{t}_{1}}x}\]

C) \[x=y\] 

D) \[\frac{x}{y}=\frac{\beta {{t}_{1}}}{\alpha {{t}_{2}}}\]

Correct Answer: A

Solution :

[a] : Slope of v - t graph = Acceleration. \[\alpha =\frac{{{v}_{0}}}{{{t}_{1}}},\beta =\frac{{{v}_{0}}}{{{t}_{2}}}\] \[\therefore \]\[\frac{\beta }{\alpha }=\frac{{{t}_{1}}}{{{t}_{1}}}\] Displacement = Area under v - t graph \[\therefore \]\[x=\frac{1}{2}{{t}_{1}}\times {{v}_{0}}\]and\[y=\frac{1}{2}{{t}_{2}}\times {{v}_{0}}\] Hence, \[\frac{x}{y}=\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\beta }{\alpha }\]