question_answer

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

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

B) \[\text{S=ft}\]

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

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

Correct Answer: D

Solution :

Distance from A to B = \[S=\frac{1}{2}f{{t}_{1}}^{{}}\] Distance from B to C = \[\left( f{{t}_{1}} \right)t\] Distance from C to D = \[\frac{{{u}^{2}}}{2a}=\frac{{{\left( f{{t}_{1}} \right)}^{2}}}{2\left( f/2 \right)}\] \[f{{t}_{1}}^{2}=2S\] \[\Rightarrow S+f{{t}_{1}}t+2S=15S\] \[\Rightarrow f{{t}_{1}}t+12S\text{                 }.....\text{(i)}\] \[\frac{1}{2}f{{t}_{1}}^{2}=S\] Dividing (i) by (ii), we get \[{{t}_{1}}=\frac{t}{6}\] \[\Rightarrow S=\frac{1}{2}f{{\left( \frac{t}{6} \right)}^{2}}=\frac{f{{t}^{2}}}{72}\]