At the instant shown, the bicyclist shown at A is traveling at 7 m/s around the curve while increasing his speed at 0.5 m/s^2. The bicyclist at B is traveling at 8.5 m/s along the straight-a-way and increasing his speed at 0.7 m/s^2. Determine the relative vrlocity and the relative acceleration of A with respect to B at this instant.(adsbygoogle = window.adsbygoogle || []).push({});

http://img108.imageshack.us/img108/2137/engr204problem122012tn.jpg [Broken]

I figured out the relative velocity:

[tex]\overrightarrow{V_B}\,=\,\left{(8.5)\,\widehat{i}\right}\,\frac{m}{s}[/tex]

[tex]\overrightarrow{V_A}\,=\,\left{(7\,cos\,50)\,\widehat{i}\,+\,(-7\,sin\,50)\,\widehat{j}\right}\,\frac{m}{s}[/tex]

[tex]\overrightarrow{V_{A/B}}\,=\,\overrightarrow{V_A}\,-\,\overrightarrow{V_B}[/tex]

[tex]\overrightarrow{V_{A/B}}\,=\,\left{[\,(7\,cos\,50)\,-\,8.5]\,\widehat{i}\,+\,(-7\,sin\,50)\,\widehat{j}\right}\,\frac{m}{s}[/tex]

[tex]\overrightarrow{V_{A/B}}\,=\,\left{(-4.00)\,\widehat{i}\,+\,(-5.36)\,\widehat{j}\right}\,\frac{m}{s}[/tex]

[tex]V_{A/B}\,=\,\sqrt{(-4.00)^2\,+\,(-5.36)^2}\,=\,6.69\,\frac{m}{s}[/tex]

The above answer for the magnitude of the velocity is correct. However, I cannot figure the acceleration part!!!

[tex](A_A)_N\,=\,\frac{V_A^2}{\rho}\,=\,\frac{7^2}{50}\,=\,0.98\,\frac{m}{s^2}[/tex]

I know the normal part of the bike at A's acceleration, but when I use this and the other EQ's for the acceleration, I don't get the correct answer! Please Help!

[tex]a_B\,=\,a_A\,+\,a_{B/A}[/tex]

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# Homework Help: Relative-Motion analysis between two particles

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