Question Number 40499 by Necxx last updated on 23/Jul/18
$${A}\:{particle}\:{of}\:{mass}\:\mathrm{2}×\mathrm{10}^{−\mathrm{27}} {kg}\:{moves} \\ $$$${according}\:{to}\:{the}\:{following} \\ $$$${y}=\mathrm{5}{cos}\left(\frac{\pi{t}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${find}\:{the}\:{maximum}\:{kinetic}\:{energy} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18
$$\frac{{dy}}{{dt}}=\mathrm{5}.{sin}\left(\frac{\Pi{t}}{\mathrm{3}}+\frac{\Pi}{\mathrm{4}}\right).\frac{\Pi}{\mathrm{3}} \\ $$$$\left(\frac{{dy}}{{dt}}\right)_{{max}} =\frac{\mathrm{5}\Pi}{\mathrm{3}} \\ $$$${k}.{E}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}×\mathrm{10}^{−\mathrm{27}} .\left(\frac{\mathrm{5}\Pi}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$
Commented by Necxx last updated on 23/Jul/18
$${Thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$$${If}\:{I}\:{decided}\:{to}\:{compute}\:{the}\:{final} \\ $$$${answer}\:{do}\:{I}\:{need}\:{to}\:{convert}\:\mathrm{5}\pi/\mathrm{3} \\ $$$${to}\:{degree}\:{before}\:{solving}\:{or}\:{running} \\ $$$${it}\:{straight}\:{just}\:{like}\:{this} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Jul/18
$${put}\:\Pi=\mathrm{3}.\mathrm{14} \\ $$