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Question Number 37648 by Rio Mike last updated on 16/Jun/18
$$\mathrm{A}\:\mathrm{particle},\:\mathrm{of}\:\mathrm{mass}\:\mathrm{5kg},\mathrm{moves}\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{its}\:\mathrm{displacement},\mathrm{x} \\ $$$$\mathrm{metres}\:\mathrm{after}\:\mathrm{t}\:\mathrm{seconds}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${x}\:=\:{t}^{\mathrm{3}} −\:\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}.\:{F}\mathrm{ind}\:\:\mathrm{the}\:\mathrm{magnitude} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{impulses}\:\mathrm{exerted}\:\mathrm{on}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{when}\:\mathrm{t}=\mathrm{2}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18
$$\frac{{dx}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{4} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }=\mathrm{6}{t}−\mathrm{8} \\ $$$${impulse}=\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{5}\left(\mathrm{6}{t}−\mathrm{8}\right){dt} \\ $$$$=\mid\mathrm{30}\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{40}{t}\mid_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\mathrm{15}×\mathrm{4}−\mathrm{80}=−\mathrm{20} \\ $$$${or} \\ $$$$\mathrm{5}\mid\mathrm{3}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{4}\mid_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\mathrm{5}\left(\mathrm{12}−\mathrm{16}+\mathrm{4}−\mathrm{4}\right)=−\mathrm{20} \\ $$$${velocity}\:{at}\:{t}=\mathrm{2}{sec} \\ $$$$\mathrm{3}×\mathrm{2}^{\mathrm{2}} −\mathrm{8}×\mathrm{2}+\mathrm{4}=\mathrm{0}\:{m}/{sec} \\ $$
Commented by Rio Mike last updated on 16/Jun/18
$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{about}}\:\boldsymbol{\mathrm{the}}\: \\ $$$$\boldsymbol{\mathrm{velocity}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{paticle}}\:\boldsymbol{\mathrm{when}} \\ $$$$\boldsymbol{\mathrm{t}}=\mathrm{2} \\ $$