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HalfThere
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Homework Statement
An object of mass M begins with a velocity of 0 m/s at a point. A power input of P watts goes directly to kinetic energy until the object has traveled a distance of X meters. What is the final velocity of the object?
So, we have constant variables
M = mass
X = distance that power will be input
P = power level
And also
V = final velocity, after traveling distance X (to be solved)
Homework Equations
E = 1/2MV[tex]^{2}[/tex]
X = [tex]\int[/tex]V(t) dt - V(t) is V as a function of time.
E = P*t
The Attempt at a Solution
Find V as a function of E (easy)
V = [tex]\sqrt{2E}[/tex]/M
Find V as a funciton of time t (use equation)
V = [tex]\sqrt{2P*t}[/tex]/M
Now take the equation for X
X = [tex]\int[/tex]V(t) dt
And Find X as a function of t directly, knowing the V(t) function
X = [tex]\int[/tex][tex]\sqrt{2P*t}[/tex]/M dt Edit: Should be, and was calculated as ([tex]\int[/tex] [tex]\sqrt{2P*t}[/tex] dt)/M
Integrate (remember that sqrt(2P) is a constant)
X(t) = (2/3)*[tex]\sqrt{2P}[/tex]*t[tex]^{3/2}[/tex]/M
Now change to t in terms of X
t(X) = ((3/2)/[tex]\sqrt{2P}[/tex]*M*X)[tex]^{2/3}[/tex]
And finally slide that into the V(t) equation
V(X) = [tex]\sqrt{2P}[/tex]*((3/2)/[tex]\sqrt{2P}[/tex]*M*X)[tex]^{1/3}[/tex]/M
Simplify (whew!)
V(X) = 3[tex]^{1/3}[/tex]*2[tex]^{-1/6}[/tex]*P[tex]^{1/3}[/tex]*X[tex]^{1/3}[/tex]*M[tex]^{-2/3}[/tex]
So, V correlates directly with the cube root of X, the cube root of P, and M^(-2/3), with a weird constant.
Am I right? Am I not? If not, where did I go wrong?
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