Deriving acceleration from velocity equation
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EDIT — (20 May 2023 at 12:46)
If this is symbolic, of course, just take the derivative with respect to t —
syms g H L t
v = sqrt(2*g*H)*tanh((sqrt((2*g*H)/(2*L)))*t)
a = diff(v,t)
.
2 个评论
Torsten
2023-5-20
My guess is that H is a function of t.
Star Strider
2023-5-20
Your guess is likely much better than mine.
In that instance, the syms call becomes:
syms g H(t) L t
and the resulting derivative difficult to work with.
However if the result is numeric, gradient would likely still work.
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2023-5-20
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