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

A particle P moves in a plane such that at time t seconds, its velocity, v=(2ti−t^3 )ms^(−1) . (a) Find, when t=2, the magnitudeof the: (i) velocity of P. (ii) acceleration of P. (b) Given that P is at the point with position vector (3i+2j) when t=1, find the position vector of P when t=2.

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

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

v=(2ti−t^3 j)ms^(−1) a. i- v_(t=2) =∣4i−8j∣=(√(4^2 +8^2 ))=4(√5)ms^(−1) ii- a(t)=v′(t)=(d/dt)(2ti−t^3 j)=(2i−3t^2 j)ms^(−2) a_(t=2) =∣2i−12j∣=(√(148))ms^(−2) =2(√(37))ms^(−2) b. P(t)=∫v(t)dt=∫(2ti−t^3 j)dt=t^2 i−(t^4 /4)j+C P(1)=3i+2j= i−(1/4)j+C⇒C=2i+(9/4)j P(t)=(t^2 +2)i−(1/4)(t^4 −9)j P(2)=6i−(7/4)j

$$\mathrm{v}=\left(\mathrm{2}{ti}−{t}^{\mathrm{3}} {j}\right)\mathrm{ms}^{−\mathrm{1}} \\ $$$${a}. \\ $$$${i}-\:\mathrm{v}_{{t}=\mathrm{2}} =\mid\mathrm{4}{i}−\mathrm{8}{j}\mid=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{5}}\mathrm{ms}^{−\mathrm{1}} \\ $$$${ii}-\:{a}\left({t}\right)=\mathrm{v}'\left({t}\right)=\frac{{d}}{{dt}}\left(\mathrm{2}{ti}−{t}^{\mathrm{3}} {j}\right)=\left(\mathrm{2}{i}−\mathrm{3}{t}^{\mathrm{2}} {j}\right)\mathrm{ms}^{−\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{a}_{{t}=\mathrm{2}} =\mid\mathrm{2}{i}−\mathrm{12}{j}\mid=\sqrt{\mathrm{148}}\mathrm{ms}^{−\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{37}}\mathrm{ms}^{−\mathrm{2}} \\ $$$$ \\ $$$${b}. \\ $$$$\mathrm{P}\left({t}\right)=\int\mathrm{v}\left({t}\right)\mathrm{dt}=\int\left(\mathrm{2}{ti}−{t}^{\mathrm{3}} {j}\right){dt}={t}^{\mathrm{2}} {i}−\frac{{t}^{\mathrm{4}} }{\mathrm{4}}{j}+\mathrm{C} \\ $$$$\mathrm{P}\left(\mathrm{1}\right)=\mathrm{3}{i}+\mathrm{2}{j}=\:{i}−\frac{\mathrm{1}}{\mathrm{4}}{j}+{C}\Rightarrow{C}=\mathrm{2}{i}+\frac{\mathrm{9}}{\mathrm{4}}{j} \\ $$$$\mathrm{P}\left({t}\right)=\left({t}^{\mathrm{2}} +\mathrm{2}\right){i}−\frac{\mathrm{1}}{\mathrm{4}}\left({t}^{\mathrm{4}} −\mathrm{9}\right){j} \\ $$$$\mathrm{P}\left(\mathrm{2}\right)=\mathrm{6}{i}−\frac{\mathrm{7}}{\mathrm{4}}{j} \\ $$

Commented by pete last updated on 09/Jul/22

Thank you sir, I am grateful.

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

Commented by Tawa11 last updated on 11/Jul/22

Great sir

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

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