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DISCUSSION THREAD
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HINTS
It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, et. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude! et = v/v .
Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration equation.
Do not confuse R with ρ - they are not the same thing.
I'm getting the rate of change of speed in Etheta. Is that correct or should I be converting that? Also would the units be ft/s^2?
For finding the rate of change of speed I am not getting my answer in e^theta. I used the equation v' = ap dot e^t, and subbed both of the variables with polar equivalents then took the dot product.
This is the way I did it too. Bouncing off of this, don't forgot that you can find e^t from (v vector)/(v magnitude) = e^t
rate of change of speed is directionless, similar to how speed is also directionless. using the dot product will give you a numerical value with unit ft/s^2 but no etheta or etangent.
I calculated radius of curvature past 700 feet is that a reasonable answer. Since I calculated speed and rate of change of speed I used the magnitude of the acceleration to find the radius of curvature. Also when calculating the magnitude of acceleration do i need to convert to path or does it not matter?
The magnitudes should be the same so I don't think it should matter.
To calculate the curvature, you need to use the magnitude of acceleration equation and solve it for p. You don't want to use a*en=v^2/p since it would be more time consuming as a result of having to solve for en.