A) \[{{f}_{0}}T\]
B) \[\frac{1}{2}\,{{f}_{0}}{{T}^{2}}\]
C) \[{{f}_{0}}{{T}^{2}}\]
D) \[\frac{1}{2}{{f}_{0}}T\]
Correct Answer: D
Solution :
Given the acceleration of the particle \[f={{f}_{0}}\,\left( 1-\frac{t}{T} \right)\] Or \[f=\frac{d\nu }{dt}={{f}_{0}}\left( 1-\frac{t}{T} \right)\] \[d\nu ={{f}_{0}}\left( 1-\frac{t}{T} \right)dt\] Integrating above equation, \[\int_{0}^{V}{dv=\,\,\int_{0}^{t}{\,{{f}_{0}}\left( 1-\frac{t}{T} \right)}}\,dt\] \[\nu \,\,=\,\,\left[ {{f}_{0}}\,t-\frac{{{f}_{0}}}{T}\cdot \,\frac{{{t}^{2}}}{2} \right]_{0}^{t}\] \[\therefore \,\,\,\,\nu ={{f}_{0}}t-\frac{{{f}_{0}}}{T}\cdot \,\,\frac{{{t}^{2}}}{2}\] If \[\operatorname{f}= 0\] Then \[f={{f}_{0}}\left( 1-\frac{t}{T} \right)=0\] \[1-\frac{t}{T}=0\,\,\,\,or\,\,\,\,\,t=T\] Substituting, \[t\text{ }=\text{ }T\] in Eq. (ii), then velocity \[{{v}_{x}}={{f}_{0}}T-\frac{{{f}_{0}}}{T}\cdot \frac{{{T}^{2}}}{2}\,\,=\,\,{{f}_{0}}T-\frac{{{f}_{0}}T}{2}=\frac{1}{2}\,\,{{f}_{0}}T\]
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