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The particle P moves along a space curve. At one instant it has velocity $v = (4i-2j-k)$ $m/s$. The magnitude of the acceleration is $8 m/s^2$. The angle between the acceleration and the velocity vector is $20^{\circ}$, so one can calculate that the acceleration in the direction of the velocity is $7.52$.

How can I calculate the radius of curvature from this information?

One of my attempts has been to try to imagine an infinitesimal change in velocity, $v = r\theta$. This implies $\frac{dv}{dt} = r\frac{d\theta}{dt}$. Could I perhaps know somehow what $\frac{d\theta}{dt}$ is?

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closed as off-topic by AccidentalFourierTransform, Kyle Kanos, John Rennie, Emilio Pisanty, Jon Custer Aug 12 '18 at 17:18

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Radius of curvature is governed by $a = v^2/r$. The radius of curvature thus calculated is good at that instant only, since 'v' will continue to increase; and, if 'a' remains constant, change 'r'.

The 'a' in the equation is the component of total acceleration which is normal to the velocity vector, or $sin(20^o)(8m/s^2)$

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The speed vector you gave is constant so there can't be any acceleration. However, for $|\textbf{v}|=u=const$, as Vintage said, $\textbf{a}=\frac{u^2}{R}\textbf{N}$, where $\textbf{N}$ is the unit vector normal to the curve. So $|\textbf{a}|=\frac{u^2}{R}|\textbf{N}|\Leftrightarrow R=\frac{u^2}{|\textbf{a}|}$.

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FYI - The formula for decomposing the acceleration vector is $ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{r} \vec{n} $ where $v$ is the speed, $\dot{v}$ the rate of change of speed (acceleration along the trajectory), the vectors $\vec{e}$ and $\vec{n}$ are along the trajectory and normal to it and $r$ is the curvature.

Look at this article for all the math the accompany this (Fenet Frames), or just look up Acceleration in wikipedia.

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