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Question-14219




Question Number 14219 by tawa tawa last updated on 29/May/17
Answered by ajfour last updated on 29/May/17
            s^� =4t^3 i^� −2t^2 j^� +3tk^�    ..(i)  (a) v^� =(ds^� /dt)=12t^2 i^� −4tj^� +3k^�    ...(ii)  (b) speed at t=1s     =∣v^� ∣_(t=1s) =(√(144+16+9)) =13  (c)  a^� =(dv^� /dt)=24ti^� −4j^�        .....(iii)  at t=3s,  a^� =72i^� −4j^�   (d) at t=0,  from eqn (ii):       v_0 ^� =3k^�    .
$$\:\:\:\:\:\:\:\:\:\:\:\:\bar {{s}}=\mathrm{4}{t}^{\mathrm{3}} \hat {{i}}−\mathrm{2}{t}^{\mathrm{2}} \hat {{j}}+\mathrm{3}{t}\hat {{k}}\:\:\:..\left({i}\right) \\ $$$$\left({a}\right)\:\bar {{v}}=\frac{{d}\bar {{s}}}{{dt}}=\mathrm{12}{t}^{\mathrm{2}} \hat {{i}}−\mathrm{4}{t}\hat {{j}}+\mathrm{3}\hat {{k}}\:\:\:…\left({ii}\right) \\ $$$$\left({b}\right)\:{speed}\:{at}\:{t}=\mathrm{1}{s} \\ $$$$\:\:\:=\mid\bar {{v}}\mid_{{t}=\mathrm{1}{s}} =\sqrt{\mathrm{144}+\mathrm{16}+\mathrm{9}}\:=\mathrm{13} \\ $$$$\left({c}\right)\:\:\bar {{a}}=\frac{{d}\bar {{v}}}{{dt}}=\mathrm{24}{t}\hat {{i}}−\mathrm{4}\hat {{j}}\:\:\:\:\:\:\:…..\left({iii}\right) \\ $$$${at}\:{t}=\mathrm{3}{s},\:\:\bar {{a}}=\mathrm{72}\hat {{i}}−\mathrm{4}\hat {{j}} \\ $$$$\left({d}\right)\:{at}\:{t}=\mathrm{0},\:\:{from}\:{eqn}\:\left({ii}\right): \\ $$$$\:\:\:\:\:\bar {{v}}_{\mathrm{0}} =\mathrm{3}\hat {{k}}\:\:\:.\: \\ $$
Commented by tawa tawa last updated on 29/May/17
God bless you sir. i really appreciate your effort.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

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