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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
 

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.
 
Dec25-12, 08:24 PM   #3
 
Quote by Pengwuino View Post
Exactly. The velocity of an object is simply the time-derivative of its position function.
Maybe I'm incredibly rusty on my calculus, but isn't the time-derivative of the first equation the following?

[tex]0={ v }_{ yi }+{ a }_{ y }t[/tex]
 
Dec25-12, 08:47 PM   #4
 
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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
 
Quote by jtbell View Post
##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

$$y(t) = y_i + v_{yi} t + \frac{1}{2}a_y t^2$$

$$v_y(t) = v_{yi} + a_y t$$
Thank you!
 
Dec25-12, 08:51 PM   #6
 
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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