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Discussion Response Post to: irodov q.....relative motion

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17 May 2008 09:45:43 IST

Subject: Re:irodov q.....relative motion

SuperCool (37)

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Using Vectors

${v}^{\rightarrow'}={v}^{\rightarrow}-{{v}_{0}}^{\rightarrow}$

Draw vector diagram,and apply triangle properties

$v'=\sqrt{{{v}_{0}}^{2}+{v}^{2}+2v{v}_{0}\phi}\\v'=39.7 Km/hr (Approx)$

And

$\frac{{v}_{0}}{Sin(\pi-\phi)}=\frac{v}{sin\theta}\\sin\theta=\frac{vsin\phi}{v'}\\\theta={Sin}^{-1}(\frac{vSin\phi}{v'})$

$\\Using\:v'\:value\\\theta={19}^{0}\:Approx.$

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