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 f/2 to come to rest. If the total distance travelled is 15 S, then       AIEEE  Solved  Paper-2005

A) \[S=ft\]

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

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

D)        \[S=\frac{1}{4}f{{t}^{2}}\]

Correct Answer: D

Solution :

The graph between velocity and time gives the distance travelled by the body in the motion. The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope and slope of   In graph, area ofgives distances,               (say) ...(i) Area of rectangle ABED gives distance travelled in time t. Distance travelled in time Thus, \[S+(f{{t}_{1}})t+ft_{1}^{2}\,=15S\] \[S+(f{{t}_{1}})t+2S=15S\]                                              ...(ii) From Eqs. (i) and (ii), we have            From Eq. (i), we get            Hence, none of the given options is correct.