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Is there some sort of calculus relationship between these two kinematics equations? 
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#1
Dec2512, 08:11 PM

P: 273

[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. 


#2
Dec2512, 08:14 PM

PF Gold
P: 7,120

Exactly. The velocity of an object is simply the timederivative of its position function.



#3
Dec2512, 08:24 PM

P: 273

[tex]0={ v }_{ yi }+{ a }_{ y }t[/tex] 


#4
Dec2512, 08:47 PM

Mentor
P: 11,778

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


#5
Dec2512, 08:48 PM

P: 273




#6
Dec2512, 08:51 PM

P: 273

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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