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A body travels a distance s in t seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f and in the second part with constant retardation r. The value of t is given by     AIEEE  Solved  Paper-2003

A)                         \[2s\left( \frac{1}{f}+\frac{1}{r} \right)\]                              

B) \[\frac{2s}{\frac{1}{f}+\frac{1}{r}}\]                        

C) \[\sqrt{2s\,(f+r)}\]      

D)       \[\sqrt{2s\,\left( \frac{1}{f}+\frac{1}{r} \right)}\]

Correct Answer: D

Solution :

Here, \[{{x}_{1}}+{{x}_{2}}=s\]                 and        \[{{t}_{1}}+{{t}_{2}}=t\] When a particle moves from A to C,                 \[{{v}^{2}}={{u}^{2}}+2f{{x}_{1}}\]                 \[{{v}^{2}}=0+2f{{x}_{1}}\] \[\Rightarrow \]               \[{{x}_{1}}=\frac{{{v}^{2}}}{2f}\]                               ... (i) and        \[v=u+ft\] \[\Rightarrow \]               \[v=0+f{{t}_{1}}\] \[\Rightarrow \]               \[{{t}_{1}}=\frac{v}{f}\]                                 ... (ii) From C to \[{{v}^{2}}={{u}^{2}}+2fs\]                 \[v{{'}^{2}}={{v}^{2}}-2r{{x}_{2}}\]                 \[0={{v}^{2}}-2r{{x}_{2}}\] \[\Rightarrow \]               \[{{x}_{2}}=\frac{{{v}^{2}}}{2r}\]                               ?. (iii) and        \[v'=v-r{{t}_{2}}\] \[\Rightarrow \]               \[0=v-r{{t}_{2}}\] \[\Rightarrow \]               \[{{t}_{2}}=\frac{v}{r}\]                                 ... (iv) On adding Eqs. (i) and (iii), we get                 \[{{x}_{1}}+{{x}_{2}}=\frac{{{v}^{2}}}{2}\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \]               \[2s={{v}^{2}}\left( \frac{1}{f}+\frac{1}{f} \right)\]           ... (v) On adding Eqs. (ii) and (iv), we get                 \[{{t}_{1}}+{{t}_{2}}=v\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \]               \[t=v\left( \frac{1}{f}+\frac{1}{r} \right)\]             \[\Rightarrow \]               \[{{t}^{2}}={{v}^{2}}{{\left( \frac{1}{f}+\frac{1}{r} \right)}^{2}}\]                ?. (vi) On dividing Eq. (vi) by Eq. (v), we get                                 \[\frac{{{t}^{2}}}{2\,s}=\frac{{{v}^{2}}{{\left( \frac{1}{f}+\frac{1}{r} \right)}^{2}}}{{{v}^{2}}\left( \frac{1}{f}+\frac{1}{r} \right)}\]                 \[\Rightarrow \]               \[\frac{{{t}^{2}}}{2\,s}=\left( \frac{1}{f}+\frac{1}{r} \right)\]                 \[\Rightarrow \]               \[t=\sqrt{2\,s\left( \frac{1}{f}+\frac{1}{r} \right)}\]