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 |
\[f={{f}_{0}}\left( 1-\frac{t}{T} \right)\] |
or \[f=\frac{dv}{dt}={{f}_{0}}\left( 1-\frac{t}{T} \right)\] \[\left[ \because \,f=\frac{dv}{dt} \right]\] |
or \[dv={{f}_{0}}\left( 1-\frac{t}{T} \right)dt\] (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 \[t=0,\text{ }v=0\] |
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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