A particle moves along a track which has a parabolic shape with a constant speed of 10m/s. The curve is given by y=5+0.3x2. Find the components of velocity and normal acceleration when x=2m.

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written 3.0 years ago by

teamques10 ★ 65k

modified 2.2 years ago by

pedsangini276 • 4.8k

engineering mechanics

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written 3.0 years ago by

teamques10 ★ 65k

$y = 5 + 0.3 x^2$ $\therefore y' = \dfrac {dy}{dx} = 0.6 x \ and \ y''=\dfrac {d^2 y}{dx^2} = 0.6$ ![](https://i.imgur.com/jndG5Kv.png) At x=2, $y' = 0.6 \times 2 = 1.2 \ and \ y''=0.6$ ∴ Radius of curvature $= \rho = \dfrac {(1+ y'^2)^{3/2}}{y''} = \dfrac {(1+1.2^2)^{3/2}}{0.6} = …

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