Question Number 16656 by Tinkutara last updated on 24/Jun/17
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{moves}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{10} \\ $$$$\mathrm{ms}^{−\mathrm{1}} \:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\:−\mathrm{2}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{direction}\:\mathrm{3}\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}.\:\mathrm{The}\:\mathrm{position}\:\mathrm{vector} \\ $$$$\mathrm{after}\:\mathrm{3}\:\mathrm{s}\:\mathrm{is} \\ $$
Answered by sma3l2996 last updated on 25/Jun/17
$${we}\:{have}\::\:\:\overset{\wedge} {{v}}={v}_{{x}} \overset{\wedge} {{i}}+{v}_{{y}} \overset{\wedge} {{j}} \\ $$$${and}\:\:{unit}\:{vector}\:{of}\:\mathrm{3}{i}+\mathrm{4}{j}\:{is}: \\ $$$$\overset{\wedge} {{e}}=\frac{\mathrm{3}}{\mathrm{5}}\overset{\wedge} {{i}}+\frac{\mathrm{4}}{\mathrm{5}}\overset{\wedge} {{j}}\: \\ $$$${so}:\:\overset{\wedge} {{v}}=\mathrm{10}.\overset{\wedge} {{e}}=\mathrm{6}\overset{\wedge} {{i}}+\mathrm{8}\overset{\wedge} {{j}} \\ $$$${and}\:{we}\:{have}\:{O}\overset{\wedge} {{M}}={x}\overset{\wedge} {{i}}+{y}\overset{\wedge} {{j}} \\ $$$$\overset{\wedge} {{v}}=\frac{{dO}\overset{\wedge} {{M}}}{{dt}}\Leftrightarrow\int{dO}\overset{\wedge} {{M}}=\int\left(\mathrm{6}\overset{\wedge} {{i}}+\mathrm{8}\overset{\wedge} {{j}}\right){dt}= \\ $$$${O}\overset{\wedge} {{M}}\left({t}\right)=\left(\mathrm{6}{t}+{c}_{{x}} \right)\overset{\wedge} {{i}}+\left(\mathrm{8}+{c}_{{y}} \right)\overset{\wedge} {{j}} \\ $$$${when}\:{t}=\mathrm{0};\:{O}\overset{\wedge} {{M}}\left({t}=\mathrm{0}\right)=\mathrm{2}\overset{\wedge} {{i}}−\mathrm{2}\overset{\wedge} {{j}}={c}_{{x}} {i}+{c}_{{y}} {j} \\ $$$${O}\overset{\wedge} {{M}}\left({t}\right)=\left(\mathrm{6}{t}+\mathrm{2}\right){i}+\left(\mathrm{8}{t}−\mathrm{2}\right){j} \\ $$$${so}\:{when}\:{t}=\mathrm{3}{s} \\ $$$${O}\overset{\wedge} {{M}}=\mathrm{20}\overset{\wedge} {{i}}+\mathrm{22}\overset{\wedge} {{j}} \\ $$$${so}\:{the}\:{position}\:{of}\:{particle}\:{after}\:\mathrm{3}{s}\:{is}:\:\:\left(\mathrm{20};\mathrm{22}\right) \\ $$
Commented by Tinkutara last updated on 25/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$