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 :
Acceleration \[f={{f}_{0}}\left( 1-\frac{t}{T} \right)\] \[orf=\frac{dv}{dt}={{f}_{0}}\left( 1-\frac{t}{T} \right)\,\left[ \because \,f=\frac{dv}{dt} \right]\] \[ordv={{f}_{0}}\left( 1-\frac{t}{T} \right)dt...(i)\] Integrating Eq. (i) on both sides, \[\int{dv=\int{{{f}_{0}}\left( 1-\frac{t}{T} \right)\,dT}}\] \[\therefore v={{f}_{0}}t-\frac{{{f}_{0}}}{T}.\frac{{{t}^{2}}}{2}+C....(ii)\] where C is constant of integration. Now, when t = 0, v = 0 So, from Eq. (ii), we get C = 0 \[\therefore v={{f}_{0}}t-\frac{{{f}_{0}}}{T}.\frac{{{t}^{2}}}{2}....(iii)\] As, \[f={{f}_{0}}\left( 1-\frac{t}{T} \right)\] When, \[f=0,\,\,0={{f}_{0}}\left( 1-\frac{t}{T} \right)\] As, \[{{f}_{0}}\ne 0\]so, \[1-\frac{t}{T}=0\,\therefore \,\,t=T\] Substituting, t = T in Eq. (iii), then velocity \[{{v}_{x}}={{f}_{0}}T-\frac{{{f}_{0}}}{T}.\frac{{{T}^{2}}}{2}\] \[={{f}_{0}}T-\frac{{{f}_{0}}T}{2}=\frac{1}{2}\,{{f}_{0}}T\]
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