question_answer

A car accelerates from rest at a constant rate \[\alpha \] for some time after which it deaccelerate at constant rate \[\beta \] and finally comes to rest. If total time elapsed is T, then maximum velocity attained by car is

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)] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{V}_{\max }}=\frac{\alpha \beta T}{(\alpha +\beta )}\]