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Question Number 71050 by jagannath19 last updated on 11/Oct/19

Commented by jagannath19 last updated on 11/Oct/19

$$explain$$

Answered by prakash jain last updated on 11/Oct/19

$$\mathrm{Given}\mathrm{that}\mathrm{graph}\mathrm{is}\mathrm{a}\mathrm{sine}\mathrm{curve},\mathrm{let}$$$$v=A\sin kt$$$$A=10fromgraph$$$$\frac{\pi }{2}=k\times 2.5\Rightarrow k=\frac{\pi }{5}$$$$v=10\sin \frac{\pi t}{5}$$$$avragevelocity=\frac{1}{5}{\int }_0^{5}10\sin \frac{\pi t}{5}$$$$=\frac{1}{5}\times 10\times {[-\frac{5}{\pi }\cos \frac{\pi t}{5}]}_9^5$$$$=\frac{20}{\pi }=\frac{20}{22}\times 7=\frac{70}{11}\approx 6.37$$

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