A) \[\frac{64}{3}m\]
B) \[\frac{64}{5}m\]
C) \[\frac{73}{3}m\]
D) \[\frac{37}{5}m\]
Correct Answer: A
Solution :
| \[\int{dv=}\int{2tdt-4}\int{dt}\Rightarrow \operatorname{v}={{t}^{2}}-4t+c;\] |
| When \[\operatorname{t}=0;\,v=0\] |
| \[c=0\]and\[V={{t}^{2}}-4t\] |
| During 0 to 4 sec motion is along -x axis. At 4 sec; velocity becomes 0 and then direction of motion reverses. |
| For finding distance we have to integrate speed and time relation. |
| Speed \[\operatorname{v}=4t-{{t}^{2}}4\sec \ge \operatorname{t}\ge 0\] |
| \[\operatorname{v}={{t}^{2}}-4\operatorname{t}\,\,\,\,\,\operatorname{t}\ge 4\sec .\] |
| \[S=\int\limits_{0}^{4}{(4t-{{t}^{2}})dt+\int\limits_{4}^{6}{({{t}^{2}}-4t)dt}}\] |
| \[=\left| 4\left( \frac{{{t}^{2}}}{2} \right)-\frac{{{t}^{3}}}{3} \right|_{0}^{4}+\left| \frac{{{t}^{3}}}{3}-\frac{4{{t}^{2}}}{2} \right|_{4}^{6}\]\[=\left| 2{{t}^{2}}-\frac{{{t}^{3}}}{3} \right|_{0}^{4}+\left| \frac{{{t}^{3}}}{3}-2{{t}^{2}} \right|_{4}^{6}\]\[=32-\frac{64}{3}+\left( \frac{216}{3}-72-\left( \frac{63}{3}-32 \right) \right)\] |
| \[=32-\frac{64}{3}+72-72-\frac{64}{3}+32=\frac{64}{3}m\] |
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