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Question Number 47194 by Necxx last updated on 06/Nov/18

The velocity of a ship Q relqtive to a ship P is 10km/h in a direction N45^. E.If the velocity of P is 20km/h in a direction N60^. W.Find the actual velocity of Q in magnitude and direction.

$${The}\:{velocity}\:{of}\:{a}\:{ship}\:{Q}\:{relqtive}\:{to} \\ $$$${a}\:{ship}\:{P}\:\:{is}\:\mathrm{10}{km}/{h}\:{in}\:{a}\:{direction} \\ $$$${N}\mathrm{45}^{.} {E}.{If}\:{the}\:{velocity}\:{of}\:{P}\:\:{is}\:\mathrm{20}{km}/{h} \\ $$$${in}\:{a}\:{direction}\:{N}\mathrm{60}^{.} {W}.{Find}\:{the} \\ $$$${actual}\:{velocity}\:{of}\:{Q}\:{in}\:{magnitude} \\ $$$${and}\:{direction}. \\ $$

Commented by Necxx last updated on 06/Nov/18

please help

$${please}\:{help} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

picture for clarity of direction...

$${picture}\:{for}\:{clarity}\:{of}\:{direction}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

V_p ^→ =−20sin60^o i+20cos60^o j velocity of Q relative to P is V_Q ^→ −V_P ^→ =10sin45^o i+10cos45^o j V_Q ^→ =i(10×(1/(√2))−20×((√3)/2))+j(10×(1/(√2))+20×(1/2)) V_Q ^→ =i(5×1.41−10×1.732)+j(5×1.41+10) =i(7.05−17.32)+j(7.05+10) =i(−10.27)+j(17.5) pls check...

$$\overset{\rightarrow} {{V}}_{{p}} =−\mathrm{20}{sin}\mathrm{60}^{{o}} {i}+\mathrm{20}{cos}\mathrm{60}^{{o}} {j} \\ $$$${velocity}\:{of}\:{Q}\:{relative}\:{to}\:{P}\:{is} \\ $$$$\overset{\rightarrow} {{V}}_{{Q}} −\overset{\rightarrow} {{V}}_{{P}} =\mathrm{10}{sin}\mathrm{45}^{{o}} {i}+\mathrm{10}{cos}\mathrm{45}^{{o}} {j} \\ $$$$\overset{\rightarrow} {{V}}_{{Q}} ={i}\left(\mathrm{10}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{20}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+{j}\left(\mathrm{10}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\mathrm{20}×\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\overset{\rightarrow} {{V}}_{{Q}} ={i}\left(\mathrm{5}×\mathrm{1}.\mathrm{41}−\mathrm{10}×\mathrm{1}.\mathrm{732}\right)+{j}\left(\mathrm{5}×\mathrm{1}.\mathrm{41}+\mathrm{10}\right) \\ $$$$\:\:={i}\left(\mathrm{7}.\mathrm{05}−\mathrm{17}.\mathrm{32}\right)+{j}\left(\mathrm{7}.\mathrm{05}+\mathrm{10}\right) \\ $$$$={i}\left(−\mathrm{10}.\mathrm{27}\right)+{j}\left(\mathrm{17}.\mathrm{5}\right) \\ $$$${pls}\:{check}... \\ $$

Commented by Necxx last updated on 06/Nov/18

yeah.... i cqlculated the modulus and direction of the value.its really correct.Thanks

$${yeah}....\:{i}\:{cqlculated}\:{the}\:{modulus} \\ $$$${and}\:{direction}\:{of}\:{the}\:{value}.{its}\:{really} \\ $$$${correct}.{Thanks} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

thanks ...

$${thanks}\:... \\ $$

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