A) \[20\sqrt{2}m\]
B) \[5m\]
C) \[15m\]
D) \[10\sqrt{2}m\]
Correct Answer: A
Solution :
| [a]: \[{{\vec{r}}_{0}}=(2.0\hat{i}+4.0\hat{j})m\] |
| \[{{\vec{v}}_{0}}=(5.0\hat{i}+4.0\hat{j})m{{s}^{-1}},\vec{a}=(4.0\hat{i}+4.0\hat{j})m{{s}^{-2}}\] |
| Along x-axis, \[{{S}_{ox}}=2m,{{V}_{ox}}=5\,m\,{{s}^{-1}}\] |
| \[{{a}_{x}}=4\,m\,{{s}^{-2}}\] |
| \[{{S}_{x}}={{S}_{ox}}+{{v}_{ox}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}\] |
| \[=2+5\times 2+\frac{1}{2}\times 4\times {{(2)}^{2}}=20m\] |
| Along y-axis, |
| \[{{S}_{oy}}=4m,{{V}_{oy}}=4m\,{{s}^{-1}},{{a}_{y}}=4m\,{{s}^{-2}}\] |
| \[{{S}_{y}}={{S}_{oy}}+{{V}_{oy}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\] |
| \[=4+4\times 2+\frac{1}{2}\times 4\times {{2}^{2}}=20m\] |
| \[S=\sqrt{S_{x}^{2}+S_{y}^{2}}=\sqrt{{{(20)}^{2}}+{{(20)}^{2}}}=20\sqrt{2}m\] |
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