question_answer

A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate \[\frac{f}{2}\] to come to rest. If the total distance traversed is 15 S, then                 [AIEEE 2005]

A)                         \[S=\frac{1}{2}f{{t}^{2}}\]                      

B)                                  \[S=\frac{1}{4}f{{t}^{2}}\]            

C)                                  \[S=\frac{1}{72}f{{t}^{2}}\]            

D)                                  \[S=\frac{1}{6}f{{t}^{2}}\]

Correct Answer: C

Solution :

                            Let car starts from point A from rest and moves up to point B with acceleration f             Velocity of car at point B,      \[v=\sqrt{2fS}\]                                                                                                   \[[\text{A}s\text{ }{{v}^{2}}={{u}^{2}}+2as]\]             Car moves distance BC with this constant velocity in time t             \[x=\sqrt{2fS}\,.\,t\]         ......(i)           [As \[s=ut\]]             So the velocity of car at point C also will be \[\sqrt{2fs}\] and finally car stops after covering distance y.             Distance CD Þ \[y=\frac{{{(\sqrt{2fS})}^{2}}}{2(f/2)}\]\[=\frac{2fS}{f}=2S\] ....(ii)                                                      \[[\text{As }{{v}^{\text{2}}}={{u}^{2}}-2as\,\Rightarrow \,s={{u}^{2}}/2a]\]             So, the total distance AD = \[AB+BC+CD\]=15S (given)             Þ            \[S+x+2S=15S\] Þ \[x=12S\] Substituting the value of x in equation (i) we get \[x=\sqrt{2fS}\,.\,t\]Þ \[12S=\sqrt{2fS}.t\] Þ \[144{{S}^{2}}=2fS.{{t}^{2}}\]             Þ \[S=\frac{1}{72}f{{t}^{2}}\].