A) 1 m/s
B) 4 m/s
C) 2 m/s
D) \[2\sqrt{2}\,m/s\]
Correct Answer: C
Solution :
From work energy theorem \[\Delta \Kappa \Epsilon =W{{E}_{net}}\] or \[{{K}_{f}}-{{K}_{i}}=\int_{{}}^{{}}{p\,dt}\] \[\frac{1}{2}m{{v}^{2}}-0=\int_{0}^{2}{\left( \frac{3}{2}{{t}^{2}} \right)}\,dt\] \[{{v}^{2}}=\left[ \frac{{{t}^{3}}}{2} \right]_{0}^{2}\] \[v=2m/s\]You need to login to perform this action.
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