A) \[\frac{1}{2}m\frac{{{\upsilon }_{1}}}{{{t}_{1}}}{{t}^{2}}\]
B) \[m\frac{{{\upsilon }_{1}}}{{{t}_{1}}}{{t}^{2}}\]
C) \[\frac{1}{2}{{\left( \frac{m{{\upsilon }_{1}}}{{{t}_{1}}} \right)}^{2}}{{t}^{2}}\]
D) \[\frac{1}{2}m{{\frac{{{\upsilon }_{1}}}{{{t}_{1}}}}^{2}}{{t}^{2}}\]
Correct Answer: D
Solution :
Work done\[=F\times S=ma\times a{{t}^{2}}=\frac{1}{2}m{{a}^{2}}{{t}^{2}}\] Now, acceleration, \[a=\] rate of change in speed\[=\frac{{{V}_{1}}}{{{t}_{1}}}\] Substituting, it in the above equation, we have work done\[=\frac{1}{2}m{{\left( \frac{{{V}_{1}}}{{{t}_{1}}} \right)}^{2}}{{t}^{2}}\]
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