question_answer

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

A) \[\left( \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta } \right)\text{t}\]     

B) \[\left( \frac{{{\alpha }^{2}}-{{\beta }^{2}}}{\alpha \beta } \right)\]

C) \[\frac{\left( {{\alpha }^{{}}}+{{\beta }^{{}}} \right)\text{t}}{\alpha \beta }\] 

D) \[\frac{\alpha \beta \text{t}}{{{\alpha }^{{}}}+{{\beta }^{{}}}}\]

Correct Answer: D

Solution :

In fig., \[\text{A}\text{.}{{\text{A}}_{\text{1}}}\text{=}{{\text{v}}_{\text{max}\text{.}}}=\alpha {{\text{t}}_{\text{1}}}=\beta {{\text{t}}_{\text{2}}}\] \[\text{But t=}{{\text{t}}_{1}}+{{\text{t}}_{2}}=\frac{{{V}_{\max }}}{\alpha }+\frac{{{V}_{\max }}}{\beta }\] \[={{\text{v}}_{\max }}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)={{\text{v}}_{\max }}\left( \frac{\alpha +\beta }{\alpha \beta } \right)\] \[or,\text{ }{{\text{v}}_{\max }}=t\left( \frac{\alpha \beta }{\alpha +\beta } \right)\]