The automobile has a speed of 27 m/s at point A and an acceleration a having a m
ID: 1705179 • Letter: T
Question
The automobile has a speed of 27 m/s at point A and an acceleration a having a magnitude of 2m/s2, acting in the direction shown. Determine the radius of curvature of the path at point A and the tangential component of acceleration
To clarify the diagram, the angle between the acceleration and the tangential axis is 30 degrees.
So the question is asking me to solve for radius and find the tangential component of acceleration at point A. I don't know how to solve for either of them, My book only demonstrates how to solve for p if either velocity is constant in which case at = 0, and a = an, which is not the case, or when the particles travel is defined by a y = f(x) equation, which also is not the case.
Please be Specific in your explanation of the solution, thank you
Explanation / Answer
angle between tangential component and resultant acceleration = 300
at = a * cos
= 2 * cos 300
= 1.732 m/s2
and the normal ( or radial) component an = a * sin
= 2 * sin 300
= 1.0 m/s2
also an = v2 / r
1.0 = 272 / r
hence the radius f path r = 272
= 729 m
tangential acceleration at = 1.732 m/s2
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