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# Answer to Question #127881 in Mechanics | Relativity for Tad Medina

Question #127881
A particle travels around a circle of radius 5 m, changing its speed at a constant rate. At a certain point A, the speed is 3m/s. After travelling another quarter revolution to point B, the speed has increased to 6 m/s. Determine the magnitude of the acceleration of the particle at B. *
1
2020-07-30T10:38:37-0400

"Solution:\\\\Find\\ the\\ tangential\\ acceleration\\\\a_\\tau=\\frac{v^2-v_0^2}{s}=\\frac{v^2-v_0^2}{0.5\\pi R}=\\\\ \n =\\frac{36-9}{0.5\\times3.14\\times5 }=3.44\\frac{m}{s^2}\\\\then\\ normal\\ acceleration\\\\a_n=\\frac{v^2}{R}=\\frac{36}{5}=7.2\\frac{m}{s^2}\\\\Acceleration\\ at\\ point\\ B\\ is \\\\the\\ vector\\ sum\\ of\\ two\\ components \\\\a=\\sqrt{a_\\tau^2+a_n^2}=\\sqrt{3.44^2+7.2^2}=7.98\\frac{m}{s^2}"

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