Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com