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 \[{{t}_{1}}=\frac{t}{6}\] \[\Rightarrow S=\frac{1}{2}f{{\left( \frac{t}{6} \right)}^{2}}=\frac{f{{t}^{2}}}{72}\]You need to login to perform this action.
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