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Question Number 175938 by ajfour last updated on 09/Sep/22

Answered by mahdipoor last updated on 09/Sep/22

$$$$$$pointofB1:$$$$(a,b)+a(sin\theta ,cos\theta )\theta =wt=\frac{v}{a}t$$$$pointofB2:$$$$(ut,0)$$$$\Rightarrow$$$$D=\mid \mid B_1{B}_2\mid {\mid }^2=$$$${(a(1+sin\theta )-ut)}^2+{(b+a.cos\theta )}^2$$$$when{\theta }_m=\alpha +\frac{\pi }{2}=\frac{2\pi }{3}\Rightarrow t_m=\frac{a\theta }{v}=\frac{2a\pi }{3v}$$$$\frac{d\theta }{dt}\left(t_m\right)=\frac{v}{a}$$$$\frac{dD}{dt}\left(t_m\right)=$$$$2(a+a.sin{\theta }_m-{ut}_m).(a.cos{\theta }_m.\frac{d\theta }{dt}\left(t_m\right)-u)+$$$$2(b+a.cos{\theta }_m)(-a.sin{\theta }_m.\frac{d\theta }{dt}\left(t_m\right))=0$$$$\Rightarrow$$$$2a(1+\frac{\sqrt{3}}{2}-\frac{2u\pi }{3v})(-\frac{v}{2}-u)+$$$$2(b-\frac{a}{2})(-\frac{v\sqrt{3}}{2})=0$$$$\Rightarrow b=\frac{a}{2}-\frac{a(1+\frac{\sqrt{3}}{2}-\frac{2u\pi }{3v})(\frac{v}{2}+u)}{\frac{v\sqrt{3}}{2}}=$$$$a\frac{\frac{2u\pi }{3v}(2u+v)-(v+u(2+\sqrt{3}))}{v\sqrt{3}}$$

Commented by ajfour last updated on 10/Sep/22

$$Thanksbothsirs.Excellent{sol}^{n}s.$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 09/Sep/22

$$\theta =\frac{vt}{a}\Rightarrow t=\frac{a\theta }{v}$$$$x_A=a+a\sin \theta$$$$y_A=b+a\cos \theta$$$$x_B=ut=\frac{ua\theta }{v}$$$$y_B=0$$$$\mathrm{\Phi }=d^2={(x_A-x_B)}^2+{(y_A-y_B)}^2$$$$\mathrm{\Phi }={(a+a\sin \theta -\frac{ua\theta }{v})}^2+{(b+a\cos \theta )}^2$$$$\frac{d\mathrm{\Phi }}{d\theta }=2(a+a\sin \theta -\frac{ua\theta }{v})(a\cos \theta -\frac{ua}{v})-2(b+a\cos \theta )\left(a\sin \theta \right)=0$$$$let\mu =\frac{b}{a},\lambda =\frac{u}{v}$$$$(1+\sin \theta -\lambda \theta )(\cos \theta -\lambda )-(\mu +\cos \theta )\sin \theta =0$$$$\Rightarrow \mu =\frac{(1+\sin \theta -\lambda \theta )(\cos \theta -\lambda )}{\sin \theta }-\cos \theta$$$$with\theta =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}$$$$\Rightarrow \mu =\frac{2(1+\frac{\sqrt{3}}{2}-\frac{2\lambda \pi }{3})(-\frac{1}{2}-\lambda )}{\sqrt{3}}+\frac{1}{2}$$$$\Rightarrow \mu =\frac{1}{2}+\frac{(4\lambda \pi -6-3\sqrt{3})(1+2\lambda )}{6\sqrt{3}}$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}.$$

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