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I thought the easiest way to approach this is to take the component of its acceleration parallel to the x-axis (as it is the point of the x-axis which will through differentiation enable us to calculate the gradient of the slope at that point).

This is what i ended up with:

a = gf`(x)/(1+f`(x)^2)

Where g= -9.81 and f`(x) is the gradient of the slope at that point.

This gives me the acceleration as a function of the position on the slope...and i have no idea how to solve this. What I would ideally like to end up with is something in the form:

x=g(t)

where "g" is any function of t. But an implicit form wouldn't be too un-elegant...