Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 55362 by Hassen_Timol last updated on 22/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Feb/19

$$plscheckthesentencemoving{30}^{o}eastetc$$

Answered by mr W last updated on 23/Feb/19

Commented by mr W last updated on 23/Feb/19

$$BmovesrelativelytoAalongpathBC$$$$withrelativespeed{v}_R.$$$$v_R=\sqrt{{(v_A+v_B)}^2{\cos }^{2}30°+{(v_A-v_B)}^2{\sin }^{2}30°}$$$$=\sqrt{{(20+10)}^2\times \frac{3}{4}+{(20-10)}^2\times \frac{1}{4}}$$$$=26.457km/h$$$$\tan \beta =\frac{(20-10)\times \sin 30°}{(20+10)\times \cos 30°}=\frac{1}{3}\times \tan 30°$$$$\Rightarrow \beta =10.893°$$$$$$$$closestdistancefromAandBis$$$$d=40\times \sin \beta =7.559km$$$$$$$$timetoreachtheclosestpointis$$$$t=\frac{40\cos \beta }{v_R}=1.485h$$

Commented by Hassen_Timol last updated on 23/Feb/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}$$$$\mathrm{God}\mathrm{bless}\mathrm{you}$$

Answered by mr W last updated on 23/Feb/19

Commented by mr W last updated on 23/Feb/19

$$anothermethod:$$$$$$$$aftertimet,$$$$x_A=v_{A}t\cos \alpha =10\sqrt{3}t$$$$y_A=v_{A}t\sin \alpha =10t$$$$x_B=40-v_{B}t\cos \alpha =40-5\sqrt{3}t$$$$y_B=v_{B}t\sin \alpha =5t$$$$distancebetweenAandB=d$$$$D=d^2={(x_A-x_B)}^2+{(y_A-y_B)}^2$$$$={(15\sqrt{3}t-40)}^2+{\left(5t\right)}^2$$$$\frac{dD}{dt}=2(15\sqrt{3}t-40)15\sqrt{3}+2\left(5t\right)5=0$$$$\Rightarrow t=\frac{6\sqrt{3}}{7}=1.485h$$$$d=\sqrt{{(15\sqrt{3}t-40)}^2+{\left(5t\right)}^2}=7.559km$$

Commented by Hassen_Timol last updated on 18/Mar/19

$$SorrySir...Canyouexplainmethemethod$$$$I^{\prime }mnotsuretohaveunderstanditentirely?$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com