# Is there some sort of calculus relationship between these two kinematics equations?

by tahayassen
Tags: calculus, equations, kinematics, relationship, sort
 P: 273 $${ y }_{ f }={ y }_{ i }+{ v }_{ yi }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\\ { v }_{ yf }={ v }_{ yi }+{ a }_{ y }t$$ It almost looks like the second equation is the derivative of the first equation with respect to time.
 PF Gold P: 7,120 Exactly. The velocity of an object is simply the time-derivative of its position function.
P: 273
 Quote by Pengwuino 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?

$$0={ v }_{ yi }+{ a }_{ y }t$$

 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 $$y(t) = y_i + v_{yi} t + \frac{1}{2}a_y t^2 \\ v_y(t) = v_{yi} + a_y t$$
P: 273
 Quote by jtbell ##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!
 P: 273 Any way I can derive the below equation from the position and velocity equations? $${ { v }_{ yf } }^{ 2 }={ { v }_{ yi } }^{ 2 }+2{ a }_{ y }({ y }_{ f }-{ y }_{ i })$$ Edit: Never mind. http://www.physicsforums.com/showthread.php?t=660863

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