Quote:
Originally Posted by Hitchhiker 42
I've attached your spreadsheets modified for 1/4*x^3 + 1
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Quote:
Originally Posted by Ether
I am assuming what you actually meant by that is
x(t) = 1/4*t^3 + 1
Is that correct?
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Quote:
Originally Posted by Hitchhiker 42
Yes, sorry about that. That was what I meant.
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Quote:
Originally Posted by Hitchhiker 42
I took the derivative of the actual velocity function. Is this not correct?
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Yes, but you got x and t confused.
given:
x(t) = 1/4*t^3 + 1
... take time derivative of x(t) to get:
x'(t) = (3/4)*t^2
...take time derivative of x'(t) to get:
x''(t) = (3/2)*t
Now solve for the differential equation:
solve x'(t) for t:
x'(t) = (3/4)*t^2 =>
t=sqrt(4x'(t)/3)
... and then substitute for t in x''(t):
x''(t) = (3/2)*t = (3/2)*
sqrt(4x'(t)/3) = sqrt(3*x'(t))