Question Number 14922 by Tinkutara last updated on 05/Jun/17
$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{particle}\:{P}\:\mathrm{due}\:\mathrm{East}\:\mathrm{is} \\ $$$$\mathrm{4}\:\mathrm{m}/\mathrm{s},\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:{Q}\:\mathrm{is}\:\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{due} \\ $$$$\mathrm{South}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:{P}\:\mathrm{w}.\mathrm{r}.\mathrm{t}. \\ $$$${Q}? \\ $$
Commented by ajfour last updated on 05/Jun/17
$$\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:{north}\:{of}\:{east}\:. \\ $$
Commented by Tinkutara last updated on 05/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}!\:\mathrm{My}\:\mathrm{little}\:\mathrm{confusion}\:\mathrm{cleared}. \\ $$
Answered by ajfour last updated on 05/Jun/17
$${take}\:{x}\:{axis}\:{along}\:{east},\:{y}\:{along} \\ $$$${south}:\:\:\:\:\:\:\bar {{v}}_{{P}} =\mathrm{4}\hat {{i}}\:\:,\:\:\:\bar {{v}}_{{Q}} =\mathrm{3}\hat {{j}} \\ $$$$\:\:\mid\bar {{v}}_{{P}} −\bar {{v}}_{{Q}} \mid=\mid\mathrm{4}\hat {{i}}−\mathrm{3}\hat {{j}}\mid\:=\sqrt{\left(\mathrm{4}\right)^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{5}\:{m}/{s}\:. \\ $$
Commented by Tinkutara last updated on 05/Jun/17
$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{velocity}? \\ $$