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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

$$a={kt}^2$$$$\frac{dv}{dt}={kt}^2$$$$dv={kt}^{2}dt$$$$v-u=\frac{{kt}^3}{3}$$$$\mathit{v}=\mathit{u}+\frac{{\mathit{k}\mathit{t}}^3}{3}$$$$\frac{dx}{dt}=u+\frac{{kt}^3}{3}$$$$dx=udt+\frac{k}{3}{t}^{3}dt$$$$\mathit{x}-{\mathit{x}}_0=\mathit{u}\mathit{t}+\frac{\mathit{k}}{12}{\mathit{t}}^4$$

Commented by chux last updated on 08/Jul/17

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

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