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Question Number 37648 by Rio Mike last updated on 16/Jun/18

$$\mathrm{A}\mathrm{particle},\mathrm{of}\mathrm{mass}5\mathrm{kg},\mathrm{moves}\mathrm{in}$$$$\mathrm{a}\mathrm{straight}\mathrm{line}\mathrm{its}\mathrm{displacement},\mathrm{x}$$$$\mathrm{metres}\mathrm{after}\mathrm{t}\mathrm{seconds}\mathrm{is}\mathrm{given}\mathrm{by}$$$$x=t^3-4t^2+4t.F\mathrm{ind}\mathrm{the}\mathrm{magnitude}$$$$\mathrm{of}\mathrm{the}\mathrm{impulses}\mathrm{exerted}\mathrm{on}\mathrm{the}\mathrm{particle}$$$$\mathrm{when}\mathrm{t}=2.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

$$\frac{dx}{dt}=3t^2-8t+4$$$$\frac{d^{2}x}{{dt}^2}=6t-8$$$$impulse={\int }_0^{2}5(6t-8)dt$$$$=\mid 30\frac{t^2}{2}-40t{\mid }_0^2$$$$=15\times 4-80=-20$$$$or$$$$5\mid 3t^2-8t+4{\mid }_0^2$$$$5(12-16+4-4)=-20$$$$velocityatt=2sec$$$$3\times 2^2-8\times 2+4=0m/sec$$

Commented by Rio Mike last updated on 16/Jun/18

$$\mathbf{how}\mathbf{about}\mathbf{the}$$$$\mathbf{velocity}\mathbf{of}\mathbf{the}\mathbf{paticle}\mathbf{when}$$$$\mathbf{t}=2$$

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