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Question Number 173293 by pete last updated on 09/Jul/22

$$\mathrm{A}\mathrm{particle}\mathbf{P}\mathrm{moves}\mathrm{in}\mathrm{a}\mathrm{plane}\mathrm{such}\mathrm{that}$$$$\mathrm{at}\mathrm{time}\mathit{t}\mathrm{seconds},\mathrm{its}\mathrm{velocity},\mathbf{v}=(2\mathrm{t}\mathit{i}-{\mathit{t}}^3){\mathrm{ms}}^{-1}.$$$$\left(\mathrm{a}\right)\mathrm{Find},\mathrm{when}t=2,\mathrm{the}\mathrm{magnitudeof}\mathrm{the}:$$$$\left(\mathrm{i}\right)\mathrm{velocity}\mathrm{of}\mathbf{P}.$$$$\left(\mathrm{ii}\right)\mathrm{acceleration}\mathrm{of}\mathbf{P}.$$$$\left(\mathrm{b}\right)\mathrm{Given}\mathrm{that}\mathbf{P}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{with}\mathrm{position}$$$$\mathrm{vector}(3\mathrm{i}+2\mathrm{j})\mathrm{when}\mathrm{t}=1,\mathrm{find}\mathrm{the}\mathrm{position}$$$$\mathrm{vector}\mathrm{of}\mathbf{P}\mathrm{when}\mathrm{t}=2.$$$$$$

Answered by Ar Brandon last updated on 09/Jul/22

$$\mathrm{v}=(2ti-t^{3}j){\mathrm{ms}}^{-1}$$$$a.$$$$i-{\mathrm{v}}_{t=2}=\mid 4i-8j\mid =\sqrt{4^2+8^2}=4\sqrt{5}{\mathrm{ms}}^{-1}$$$$ii-a\left(t\right)={\mathrm{v}}^{\prime }\left(t\right)=\frac{d}{dt}(2ti-t^{3}j)=(2i-3t^{2}j){\mathrm{ms}}^{-2}$$$$a_{t=2}=\mid 2i-12j\mid =\sqrt{148}{\mathrm{ms}}^{-2}=2\sqrt{37}{\mathrm{ms}}^{-2}$$$$$$$$b.$$$$\mathrm{P}\left(t\right)=\int \mathrm{v}\left(t\right)\mathrm{dt}=\int (2ti-t^{3}j)dt=t^{2}i-\frac{t^4}{4}j+\mathrm{C}$$$$\mathrm{P}\left(1\right)=3i+2j=i-\frac{1}{4}j+C\Rightarrow C=2i+\frac{9}{4}j$$$$\mathrm{P}\left(t\right)=(t^2+2)i-\frac{1}{4}(t^4-9)j$$$$\mathrm{P}\left(2\right)=6i-\frac{7}{4}j$$

Commented by pete last updated on 09/Jul/22

$$\mathrm{Thank}\mathrm{you}\mathrm{sir},\mathrm{I}\mathrm{am}\mathrm{grateful}.$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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