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
![Given that graph is a sine curve, let v=Asin kt A=10 from graph (π/2)=k×2.5 ⇒k=(π/5) v=10sin ((πt)/5) avrage velocity=(1/5)∫_0 ^5 10sin ((πt)/5) =(1/5)×10×[−(5/π)cos ((πt)/5)]_9 ^5 =((20)/π)=((20)/(22))×7=((70)/(11))≈6.37](https://www.tinkutara.com/question/Q71070.png)
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{graph}\:\mathrm{is}\:\mathrm{a}\:\mathrm{sine}\:\mathrm{curve},\:\mathrm{let} \\ $$$${v}={A}\mathrm{sin}\:{kt} \\ $$$${A}=\mathrm{10}\:{from}\:{graph} \\ $$$$\frac{\pi}{\mathrm{2}}={k}×\mathrm{2}.\mathrm{5}\:\Rightarrow{k}=\frac{\pi}{\mathrm{5}} \\ $$$${v}=\mathrm{10sin}\:\frac{\pi{t}}{\mathrm{5}} \\ $$$${avrage}\:{velocity}=\frac{\mathrm{1}}{\mathrm{5}}\int_{\mathrm{0}} ^{\mathrm{5}} \mathrm{10sin}\:\frac{\pi{t}}{\mathrm{5}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}×\mathrm{10}×\left[−\frac{\mathrm{5}}{\pi}\mathrm{cos}\:\frac{\pi{t}}{\mathrm{5}}\right]_{\mathrm{9}} ^{\mathrm{5}} \\ $$$$=\frac{\mathrm{20}}{\pi}=\frac{\mathrm{20}}{\mathrm{22}}×\mathrm{7}=\frac{\mathrm{70}}{\mathrm{11}}\approx\mathrm{6}.\mathrm{37} \\ $$