Question Number 62889 by Tawa1 last updated on 26/Jun/19
Answered by Kunal12588 last updated on 26/Jun/19
$$\left({a}\right){s}={t}^{\mathrm{2}} −\mathrm{5}{t}+\mathrm{6} \\ $$$${at}\:{initial}\:{pos}.\:{t}=\mathrm{0} \\ $$$${s}\left(\mathrm{0}\right)=\mathrm{6} \\ $$$$\therefore{the}\:{initial}\:{dist}.\:{of}\:{particle}\:{from}\:{O}\:{is}\:\mathrm{6}{m} \\ $$$$\left({b}\right)\:{s}\left(\mathrm{3}\right)=\mathrm{9}−\mathrm{15}+\mathrm{6}=\mathrm{0} \\ $$$$\because\:{motion}\:{is}\:{in}\:{straight}\:{line} \\ $$$${distance}=\mid{initial}\:{pos}.−{final}\:{pos}.\mid \\ $$$$=\mid\mathrm{6}−\mathrm{0}\mid \\ $$$$=\mathrm{6}{m} \\ $$
Commented by Tawa1 last updated on 26/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$