Integration Questions

Question Number 17599 by chux last updated on 08/Jul/17

$$\mathrm{a}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial} \\$$$$\mathrm{speed}\:\mathrm{u},\mathrm{it}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\$$$$\mathrm{line}\:\mathrm{with}\:\mathrm{an}\:\mathrm{accleration}\:\mathrm{which} \\$$$$\mathrm{varies}\:\mathrm{as}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{time} \\$$$$\mathrm{the}\:\mathrm{particle}\:\mathrm{has}\:\mathrm{been}\:\mathrm{in}\:\mathrm{motion}. \\$$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{at}\:\mathrm{any}\:\mathrm{time}\:\mathrm{t},\mathrm{and} \\$$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}. \\$$

Answered by Tinkutara last updated on 08/Jul/17

$${a}\:=\:{kt}^{\mathrm{2}} \\$$$$\frac{{dv}}{{dt}}\:=\:{kt}^{\mathrm{2}} \\$$$${dv}\:=\:{kt}^{\mathrm{2}} {dt} \\$$$${v}\:−\:{u}\:=\:\frac{{kt}^{\mathrm{3}} }{\mathrm{3}} \\$$$$\boldsymbol{{v}}\:=\:\boldsymbol{{u}}\:+\:\frac{\boldsymbol{{kt}}^{\mathrm{3}} }{\mathrm{3}} \\$$$$\frac{{dx}}{{dt}}\:=\:{u}\:+\:\frac{{kt}^{\mathrm{3}} }{\mathrm{3}} \\$$$${dx}\:=\:{udt}\:+\:\frac{{k}}{\mathrm{3}}{t}^{\mathrm{3}} {dt} \\$$$$\boldsymbol{{x}}\:−\:\boldsymbol{{x}}_{\mathrm{0}} \:=\:\boldsymbol{{ut}}\:+\:\frac{\boldsymbol{{k}}}{\mathrm{12}}\boldsymbol{{t}}^{\mathrm{4}} \\$$

Commented by chux last updated on 08/Jul/17

$$\mathrm{wow}.....\:\mathrm{thankz} \\$$