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Is there some sort of calculus relationship between these two kinematics equations? |
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| Dec25-12, 08:11 PM | #1 |
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Is there some sort of calculus relationship between these two kinematics equations?
[tex]{ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t[/tex]
It almost looks like the second equation is the derivative of the first equation with respect to time. |
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| Dec25-12, 08:14 PM | #2 |
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Exactly. The velocity of an object is simply the time-derivative of its position function.
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| Dec25-12, 08:24 PM | #3 |
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[tex]0={ v }_{ yi }+{ a }_{ y }t[/tex] |
| Dec25-12, 08:47 PM | #4 |
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Mentor
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Is there some sort of calculus relationship between these two kinematics equations?
##v_{yf}## isn't a constant, it's a variable, more specifically the dependent variable, a function of t. Written as functions, your two equations are
$$y(t) = y_i + v_{yi} t + \frac{1}{2}a_y t^2 \\ v_y(t) = v_{yi} + a_y t$$ |
| Dec25-12, 08:48 PM | #5 |
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| Dec25-12, 08:51 PM | #6 |
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Any way I can derive the below equation from the position and velocity equations?
[tex]{ { v }_{ yf } }^{ 2 }={ { v }_{ yi } }^{ 2 }+2{ a }_{ y }({ y }_{ f }-{ y }_{ i })[/tex] Edit: Never mind. http://www.physicsforums.com/showthread.php?t=660863 |
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