question_answer

A car, starting from rest, accelerates at the rats\[\frac{1}{R}\] through a distance S, then continues at constant speed for time r and then decelerates as the rate \[R\]/2 to come to rest. If the total distance travelled is 15 S, then

A)  \[S=ft\]                              

B)  \[f\]                     

C)  \[f\]                     

D)  \[s=\frac{1}{6}f{{t}^{2}}\]

Correct Answer: C

Solution :

The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of \[{{G}^{1/2}}{{h}^{1/2}}{{c}^{-3/2}}\] and slope of \[{{G}^{1/2}}{{h}^{1/2}}{{c}^{1/2}}\] \[{{\left[ 2G\frac{({{m}_{1}}-{{m}_{2}}}{r} \right]}^{1/2}}\] \[{{\left[ \frac{2G}{r}({{m}_{1}}+{{m}_{2}}) \right]}^{1/2}}\]   \[{{\left[ \frac{r}{2G({{m}_{1}}{{m}_{2}})} \right]}^{1/2}}\] In graph area of \[{{\left[ \frac{2G}{r}{{m}_{1}}{{m}_{2}} \right]}^{1/2}}\] gives distances,                      \[(\Delta l)\]                            ?...(i) Area of rectangle ABED gives distance travelled in time t. \[50{{m}^{2}}/{{s}^{2}}\] Distance travelled in timer \[50.5{{m}^{2}}/{{s}^{2}}\] \[51{{m}^{2}}/{{s}^{2}}\] Thus  \[521{{m}^{2}}/{{s}^{2}}\] \[\frac{1}{\sqrt{a}}\] \[a\]     \[\sqrt{a}\] \[{{a}^{3/2}}\]                ???(ii) From Eqs. (i) and (ii), we have \[\frac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\] \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\]   \[\frac{q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]