A) \[\frac{\alpha +\beta }{\alpha \beta }T\]
B) \[\frac{\alpha \beta }{\alpha +\beta }T\]
C) \[\frac{\alpha }{\alpha +\beta }T\]
D) \[\frac{\beta }{\alpha +\beta }T\]
Correct Answer: B
Solution :
Let car accelerates for time \[{{T}_{1}}\] and decelerates for time \[{{T}_{2}}\] So, \[{{\operatorname{T}}_{1}}+{{T}_{2}}=T\] The corresponding velocity time graph is shown in figure From graph, \[\alpha = slope of line,\,\,OM=\frac{{{V}_{\max }}}{{{T}_{1}}}\,\,\Rightarrow \,\,{{T}_{1}}=\frac{{{V}_{\max }}}{\alpha }\] and \[\beta \,\,=\,\,-\,(slope\,\,of\,\,line\,\,MN)\,\,=\,\,\frac{{{V}_{\max }}}{{{T}_{2}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{T}_{2}}=\frac{{{V}_{\max }}}{\beta }\] Now, \[{{\operatorname{T}}_{1}}\,\,+\,\,{{T}_{2}}=T\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{V}_{\max }}}{\alpha }+\frac{{{V}_{\max }}}{\beta }=T\] [from Eqs. (i) and (ii)] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{V}_{\max }}=\frac{\alpha \beta T}{(\alpha +\beta )}\]You need to login to perform this action.
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